.u^ ; 


ivHiiC"::';;''':^'^.'"/' 


SECOND-YEAR  MATHEMATICS 
for  SECONDARY  SCHOOLS 


4 


THE  UNIVERSITY  OP  CHICAGO  PRESS 
CHICAGO,  ILLINOIS 


THE  BAKER  &  TAYLOR  COMPANY 

NEW  TOBK 

THE  CAMBRIDGE  UNIVERSITY  PRESS 

LONDON  AND  EDIKBUBSH 

THE  MARUZEN-KABUSHIKI-KAISHA 

TOKTO,  OSAKA,    KTOTO,   FCKUOKA,   SBNDAI 

THE  MISSION  BOOK  COMPANY 

SHANeHAI 


FELIX  KLEIN 

FELIX  KLEIN,  probably  the  most  eminent  German 
mathematician  of  his  time,  was  born  at  Diisseldorf  in 
1849  and  is  still  living,  though  he  retired  from  profes- 
sional life  in  1912.  He  studied  at  Bonn,  Germany,  where 
at  the  age  of  seventeen  he  became  assistant  to  the  re- 
nowned physicist,  Pliicker,  in  the  Physical  Institute.  He 
took  his  Doctor's  degree  at  eighteen  years  of  age,  then 
went  to  BerHn,  and  a  Uttle  later  to  Gottingen.  Here  he 
assisted  in  editing  Pliicker's  works. 

Klein  entered  the  Gottingen  university  faculty  in  1871. 
The  next  year  he  became  professor  of  mathematics  at 
Erlangen,  and  afterward  held  professorships  at  the  Tech- 
nical Institute  of  Munich  (1875-80)  and  at  the  universi- 
ties of  Leipzig  (1880-86)  and  of  Gottingen  (1886-1912). 
He  was  sent  to  the  World's  Fair  at  Chicago  in  1893  by 
the  Prussian  government,  to  represent  the  university 
interests  o!  the  nation. 

Klein's  pupils  are  found  in  most  of  the  leading  univer- 
sities of  the  United  States.  No  one  else  in  'Germany  has 
exerted  so  great  an  influence  on  American  mathematics. 
He  has  been  a  tireless  worker  himself,  both  in  the  science 
and  in  the  improvement  of  the  teaching  of  mathematics. 
He  was  made  president  of  the  International  Commission 
on  the  Teaching  of  Mathematics,  in  1908,  by  the  Fourth 
International  Congress  of  Mathematicians  held  that  year 
in  Rome,  Italy. 

His  contributions  to  mathematics  are  extensive,  but 
they  cannot  even  be  enumerated  here.  It  is  scarcely  too 
much  to  assert  that  Klein  has  led  the  main  movements  for 
advancement  of  mathematical  teaching  since  the  begin- 
ning of  the  present  century.  This  appUes  not  only  to 
university  teaching  but  to  secondary  (high-school) 
teaching  as  well. 

In  his  Teaching  of  Geometry,  p.  69,  Professor  David 
Eugene  Smith  of  Teachers'  College,  Columbia  University, 
says  of  Klein:  "He  has  the  good  sense  to  look  at  some- 
thing besides  good  mathematics:  (1)  he  insists  upon  the 
psychological  point  of  view;  (2)  he  demands  careful 
selection  of  subject-matter;  (3)  he  insists  on  reasonable 
correlation  with  practical  work;  (4)  he  looks  with  favor 
upon  the  union  of  plane  and  soUd  geometry;  (5)  he 
favors  also  the  union  of  algebra  and  geometry." 

Some  of  Klein's  best  interpreters  have  said  of  his 
reformatory  movement  that  Klein's  main  idea  is  to  make 
"functional  thinking  in  its  geometrical  form"  the  dis- 
tinguishing mark  of  secondary-school  work  in  mathematics. 


/T/^ 


Second-iear  Mathematics 
for   Secondary    ochools 


BY 

ERNST  R.  BRESLICH 

Head  of  the  Department  of  Alathematics  in  the  University 
High  School,  The  University  of  Chicago 


THE  UNIVERSITY  OF  CHICAGO  PRESS 
CHICAGO,  ILLINOIS 


Copyright  igio  and  1916  By 
The  University  op  Chicago 


All  Rights  Reserved 


Published  June  1910 

Second  Impression  October  191  o 

Third  Impression  August  191 1 

Fourth  Impression  October  191 2 

Fifth  Impression  October  1913 

Second  Edition  August  1916 

Second  Impression  September  1918 

Third  Impression  November  191 8 

Fourth  Impression  October  191 9 


Composed  and  Printed  By 

The  University  of  Chica?o  Press 

Chicago,  Illinois,  U.S.A. 


EDITOR'S  PREFACE 

This  book  by  its  copyright  purports  to  be  a  second 
edition  of  a  former  text  of  the  same  title  by  other  authors. 
It  is  this  in  the  sense  that  it  carries  forward  through  the 
second  high-school  year  a  reconstructed  form  of  the  union 
mathematics  of  a  first-year  text.  It  allies  itself  with  the 
former  work  also  in  that  it  places  chief  emphasis  on  plane 
geometry. 

The  older  Second-Year  Mathematics  was  an  attempt 
to  furnish  a  concrete  contribution  to  the  problem  of 
introducing  greater  homogeneity  and  continuity  into 
the  secondary  mathematical  subjects  from  year  to  year. 
In  this  particular  also  this  book  resembles  the  earlier 
text. 

In  a  very  real  sense,  however,  this  volume  is  a  new  con- 
tribution, with  its  own  plans  and  purposes.  Its  primal 
aim  is  to  furnish  a  gradually  progressive  continuation  of 
the  form  of  reconstructed  mathematics  of  the  text  First- 
Year  Mathematics,  by  Mr.  Breslich  himself.  It  aims 
definitely  to  teach  how  to  study  pbS  well  as  the  content  of  a 
second  unit  of  secondary  mathematics.  It  accomplishes 
this  through  the  nature  and  form  of  the  material,  through 
explicit  exhibits  and  formulated  tests  of  sound  and  un- 
sound reasoning,  through  study-helps,  directions  for  work- 
ing, and  systematic  chapter  summaries. 

It  seeks  neither  to  eliminate  nor  to  curtail  inherent 
mathematical  formalism,  but  to  fill  forms  and  technique 
with  the  meanings  that  flow  from  a  well-balanced  treat- 
ment of  related  material  drawn  from  the  kindred  ele- 
mentary subjects.     A  unique  feature  of  the  book  is  the 


206.1^:^ 


viii  EDITOR'S  PREFACE 

attractive  presentation  of  a  considerable  body  of  associ- 
ated solid  geometry.  This  is  an  economy  and  is  in  ac- 
cord with  modern  educational  precepts. 

The  cordial  response  from  the  best  sources  that  Mr. 
Breslich's  First-Year  Mathematics  has  met  in  the  first 
year  after  its  publication  proves  his  first  text  to  be 
generally  adaptable  to  classroom  conditions,  and  augurs 
that  the  present  book  will  be  found  to  work  smoothly 
under  average  conditions.  An  examination  of  the  con- 
text is  sufficient  to  convince  the  open-minded  reader 
that  the  educational  results  of  this  book  will  greatly 
surpass  those  of  the  text  it  is  displacing,  as  well  as  those  of 
any  standard  text  treating  plane  geometry  as  a  separate 
subject. 

G.  W.  Myers 

Chicago,  III. 

August,  1916 


AUTHOR'S  PREFACE 

In  planning  the  work  of  the  second  year  the  author 
has  kept  in  mind  the  following  facts : 

1.  Through  the  second  year  the  combined  type  of  material 
of  the  mathematics  taught  in  the  first  year  is  to  he  carried 
forward,  the  emphasis  here  being  shifted  to  geometry, 

2.  The  operations  and  laws  of  arithmetic  are  to  be  reviewed 
wherever  opportunity  is  offered  or  occasion  warrants,  as  in 
the  evolution  of  formulas,  in  the  introduction  of  new 
algebraic  topics,  and  in  problems  of  calculation. 

3.  The  algebraic  ground  gained  in  the  first  year  is  to  be 
held  and  the  field  extended  at  least  as  far  as  is  customary 
with  the  algebra  before  the  third  year. 

A  firm  hold  is  kept  on  algebra  by  the  employment  of 
algebraic  notation  and  by  the  continued  application  of 
the  equation  to  geometrical  matters.  New  algebraic 
topics  are  developed  when  opportunity  and  need  arise. 
Thus,  elimination  by  comparison  and  by  substitution,  so 
frequently  needed  in  proofs  and  in  the  solution  of  exer- 
cises, is  taught  very  early.  The  solution  of  the  quad- 
ratic equation  by  means  of  the  formula,  the  operations 
with  fractions,  and  factoring  are  all  reviews  or  further 
extensions  of  topics  begun  in  the  first  year. 

4.  The  study  of  plane  geometry  is  to  be  completed. 

In  the  first-year  course  the  student  has  gained  a  thor- 
ough understanding  of  the  fundamental  notions  of 
geometry.  Accordingly,  in  the  second  year,  methods  of 
a  more  formal  character  are  introduced  from  the  start. 
But  even  before  this,  the  advantage  of  the  reasoning 
process  over  the  process  of  measuring  has  been  recognized. 


X  AUTHOR'S  PREFACE 

Mathematical  fallacies  and  optical  deceptions  are  now 
used  to  make  the  need  of  logical  proof  still  more  .apparent. 

A  definite  aim  is  to  give  the  pupil  something  of  the 
secret  of  geometrical  strategy,  i.e.,  some  skill  in  attacking, 
taking  possession  of,  and  exploiting  a  geometrical  diffi- 
culty. With  this  in  view,  methods  of  proof  are  discussed 
and  emphasized)  not  once  for  all,  but  throughout  the 
course. 

To  cultivate  versatiUty  and  system,  students  are  taught 
to  choose  between  various  methods  of  proof,  and  always 
to  follow  some  definite  plan  and  not  to  trust  to  the  chance 
of  stumbling  upon  a  proof.  To  this  end  many  model 
proofs  are  given.  With  other  proofs  statements  or  rea- 
sons that  are  more  or  less  apparent  to  the  student  are 
omitted,  in  order  that  he  may  also  acquire  the  habit  of 
independent  thought  and  that  his  powers  of  argumen- 
tation may  be  strengthened. 

The  custom  of  dividing  the  subject  of  geometry  into  a 
few  ''books"  has  been  abandoned  as  being  of  only  tra- 
ditional or  historical  value.  The  course  is,  however, 
divided  into  a  number  of  short  chapters,  each  dealing  with 
one  or  but  a  few  central  topics.  This  arrangement  is  far 
better  adapted  to  the  study  of  high-school  students  than 
is  the  traditional  grouping  into  ''books,"  since  the  aims 
and  purposes  of  the  short  chapters  are  easily  seen.  It  is 
found  to  be  more  economical  of  the  student's  time  and 
energy  than  the  old  method. 

5.  The  student  should  receive  training  in  both  plane  and 
solid  geometry. 

Many  theorems  of  solid  geometry  that  are  closely 
related  to  corresponding  theorems  in  plane  geometry 
are  proved  in  the  second  year,  thus  furnishing  the 
student  appropriate  exercise  in  both  two-  and  three- 
dimensional  thinking. 


AUTHOR'S  PREFACE  xi 

A  real  advance  is  thus  made  in  the  study  of  solid 
geometry  before  plane  geometry  is  completed.  The  work 
in  solid  geometry  includes  the  theorems  on  lines  and  planes 
in  space  and  on  diedral  angles. 

6.  The  study  of  trigonometry  begun  in  the  first  year  is  to 
be  continued. 

It  is  a  distinct  educational  loss  that  the  strong  appeal 
that  trigonometry  has  for  high-school  pupils  should  not 
be  utilized  earlier  than  is  customary.  Moreover,  trigo- 
nometric methods  here  often  replace  algebraic  and  geo- 
metric methods,  giving  the  student  the  opportunity  to 
see  some  of  the  advantages  of  trigonometry  over  algebra 
and  geometry. 

In  addition  to  the  foregoing  aims  the  following  are 
included:  (a)  the  application  of  three  trigonometric 
functions  (sine,  cosine,  and  tangent)  to  the  solution  of 
the  right  triangle  and  to  a  number  of  practical  problems; 
(6)  the  development  of  some  of  the  fundamental  relations 
between  these  important  functions. 

7.  No  topical  treatment  of  the  theory  of  limits  is 
intended. 

Such  a  treatment  is  believed  not  to  belong  to  the  early 
years  of  the  high-school  course.  However,  the  question 
of  the  existence  of  incommensurable  lines  and  numbers  is 
raised,  examples  of  these  are  given,  and  the  notion  of  the 
limit  of  a  sequence  is  developed. 

8.  Since  the  usefulness  of  a  study  always  appeals  very 
strongly  to  a  beginner,  this  phase  is  emphasized  throughout 
the  course. 

The  importance  and  the  significance  of  geometrical 
facts  in  the  affairs  of  everyday  life  are  impressed  upon 
the  pupil.  This  wins  his  sanction  of  the  worth  of  the 
study  to-  himself  more  fully  than  any  other  sort  of 
appeal  that  the  teacher  of  geometry  can  make. 


xii  AUTHOR'S  PREFACE 

9.  The  plan  of  introducing  definitions  whenever  needed 
and  not  before,  which  is  used  in  the  first-year  course,  has  been 
followed  also  in  the  second  year. 

After  definitions  are  introduced  they  are  continually 
used,  in  order  that  the  pupil  may  acquire  mastery 
through   use. 

The  material  as  arranged  in  this  course  opens  to  the 
student  a  broader,  richer,  more  useful,  and  therefore  more 
alluring  field  of  ideas,  and  lays  a  more  stable  foundation 
for  future  work,  than  does  any  separate  treatment.  A 
great  saving  of  the  student's  time  is  effected  by  developing 
arithmetic,  algebra,  geometry,  and  trigonometry  side 
by  side.  This  union  of  subjects  also  makes  unnecessary 
the  long  and  tiresome  reviews  usually  given  at  the  begin- 
ning of  each  subject,  and  gives  place  for  frequent  inci- 
dental reviews  leading  immediately  to  an  extension  of  the 
subject. 

Often  a  high-school  pupil  fails  rightly  to  esteem  a 
high-school  subject  because  he  cannot  discern  its  bearing 
either  on  what  has  preceded  or  on  what  is  to  follow. 
But,  having  experienced  the  closeness  of  the  relation 
between  the  subjects  he  does  not  lose  sight  of  the 
familiar  fields  while  he  is  obtaining  an  outlook  into 
neighboring  and  more  remote  ones.  There  is  thus  an 
economy  resulting  both  from  accompHshing  more  work 
in  less  time  and  from  the  performance  of  tasks  that  are 
intelligently  motivated. 

The  book  contains  exercises  in  sufficiently  large  num- 
bers to  allow  the  instructor  some  choice  in  case  he  wishes 
to  reduce  the  scope  of  the  course.  Problems  and  theorems 
which  may  be  omitted  are  marked  with  the  symbol  J. 
These  problems  may  be  taken  either  in  the  course  by  the 
stronger  pupils  or  at  the  end  of  the  course  by  all.    If 


AUTHOR'S  PREFACE  xiii 

taken  at  the  end  of  the  course,  they  will  give  the  stu- 
dent ample  drill  and  review  of  the  right  sort. 

*' Second-Year  Mathematics^^  may  be  used  successfully 
in  classes  that  have  had  only  algebra  during  the  first  year. 

The  syllabus  at  the  beginning  of  the  book  gives  all  the 
theorems  and  axioms  taught  in  First-Year  Mathematics, 
indicating  the  order  in  which  they  were  given.  This 
furnishes  an  effective  introduction  to  the  formal  geometry 
of  the  second  year,  especially  so  if  it  is  taught  by  the 
syllabus  method.  It  helps  the  student  very  materially 
in  overcoming  the  difficulties  usually  encountered  in 
beginning  demonstrative  geometry,  and  at  the  same  time 
it  gives  him  the  opportunity  of  availing  himself  of  all 
the  advantages  of  the  correlation  of  algebra,  geometry, 
and  trigonometry  in  the  second  year. 

The  author  desires  to  render  acknowledgment  to 
Professor  Charles  H.  Judd  for  his  numerous  suggestions 
and  criticisms.  His  recent  book  on  The  Psychology  of 
High-School  Subjects  has  been  of  invaluable  service  in 
planning  this  course. 

The  encouragement,  interest,  and  advice  of  Principal 
F.  W.  Johnson,  of  the  University  High  School,  have 
been  a  very  substantial  help  in  bringing  about  the  publica- 
tion of  this  course. 

The  author  is  also  indebted  to  his  colleagues,  Messrs 
Raleigh  Schorling,  Horace  C.  Wright,  and  Harry  N 
Irwin,  who  have  read  and  criticized  in  detail  every 
chapter  of  the  book. 

The  portraits  of  Fermat  and  Gauss  which  are  used  as 
inserts  in  the  text  have  been  taken  from  the  "Philo- 
sophical Portrait  Series,"  pubHshed  by  the  Open  Court 
Publishing  Company,  Chicago. 

Ernst  R.  Breslich 
Chicago,  III. 
September,  1916 


CONTENTS 

CHAPTEB  PAGE 

I.  Assumptions,  Theorems,  and  Construc- 
tions Given  in  First-Year  Mathe- 
matics       1 

II.  Methods  of  Proof 8 

Logic "  .     .     .     .  8 

Geometrical  Fallacies 9 

Need  for  Proof . 11 

Methods  of  Proof 12 

III.  Methods  of  Elimination,  Problems  and 

Exercises  in  Two  Unknowns    ...  23 

IV.  Quadrilaterals.      Prismatic   Surface. 

DiEDRAL  Angles 32 

Parallelograms 32 

Quadratic  Equations 49 

The  Trapezoid 49 

The  Kite 49 

Symmetry 50 

Loci 51 

Surfaces       .      .      .    ^ 51 

Lines  and  Planes  in  Space 53 

Diedral  Angles 56 

V.  Proportional  Line-Segments  ....  59 

Uses  of  Proportional  Line-Segments      .      .  59 

Proportional  Segments 62 

Problems  of  Construction 73 

Lines  and  Planes  in  Space 76 

VI.  Proportion.    Factoring.     Variation     .  81 

Fundamental  Theorems  of  Proportion  .      .  81 

Factoring 84 

XV 


XVI  CONTENTS 

CHAPTER  PAGE 

Proportions  Obtained  from  Given  Propor- 
tions    87 

Relation  between  Proportion  and  Variation  95 

VII.  Similar  Polygons 100 

Uses  of  Similar  Triangles 100 

Theorems  on  Similar  Figures       ....  106 

VIII.  Relations  between  the  Sides  of  Tri- 
angles. Theorem  of  Pythagoras  and 
Its  Generalizations.  Quadratic  Equa- 
tions.    Radicals 116 

Similarity  in  the  Right  Triangle       .      .      .  116 

Radicals 118 

Problems  of  Construction 120 

Relations  of  the  Sides  of  a  Right  Triangle  122 

Quadratic  Equations 125 

The    Generalization    of    the    Theorem    of 

Pythagoras 132 

IX.  Trigonometric  Ratios.    Radicals.    Quad- 
ratic Equations  in  Two  Unknowns     .  137 

Trigonometric  Ratios 137 

Exact  Values  of  the  Functions  of  30°,  45°, 

and  60° 142 

Radicals 143 

Application  of  the  Trigonometric  Functions  145 

Relations  of  Trigonometric  Functions   .      .  152 
Quadratic  Equations  Solved  by  Graph  and 

by  Elimination 155 

X.  The  Circle 160 

•Review  and  Extension  of  the  Properties  of 

the  Circle 160 

Tangent  Circles 168 


CONTENTS  xvii 

CHAPTER  PAGE 

XI.  Measurement  of  Angles  by  Arcs  of  the 

Circle 173 

Units  of  Angular  Measure 173 

Inscribed  Angles 174 

Problems  of  Construction 182 

XII.  Proportional  Line-Segments  in  Circles  194 

XIII.  Operations  WITH  Fractions.    Fractional 

Equations 201 

Addition  and  Subtraction  of  Fractions       .  201 

Multiplication  of  Fractions 205 

Division  of  Fractions 208 

Complex  Fractions 210 

Fractional  Equations 212 

Trigonometric  Relations 217 

XIV.  Inequalities 220 

Review  and  Extension  of  the  Axioms  and 

Theorems  of  Inequality  Previously  Estab- 
lished         220 

Solution   of   Problems   by  Means  of   In- 
equalities         223 

Theorems  of  Inequality 226 

Lines  and  Planes  in  Space 238 

XV.  Lines  and  Planes  in  Space.     Diedral 

Angles.    The  Sphere 243 

Lines  and  Planes  in  Space 243 

Diedral  Angles       .      .      .      .      .      .      .      .251 

The  Sphere 255 

XVI.  Loci.    Concurrent  Lines 264 

Loci 264 

Concurrent  Lines 270 


xviil  CONTENTS 


PAGE 


XVII.  Regular   Polygons   Inscribed   in,    and 
Circumscribed     about,    the     Circle. 

Length  of  the  Circle 279 

Construction  of  Regular  Polygons    .      .      .  279 

The  Length  of  the  Circle 294 

XVIIL  Comparison  of  Areas.  Literal  Equa- 
tions. Area  of  the  Triangle.  Fac- 
toring       302 

Comparison  of  Areas 302 

Literal  Equations  in  One  Unknown       .      .  304 
Systems    of    Linear    Equations    in    Two 

Unknowns  Having  Literal  Coefficients    .  306 

The  Area  of  the  Triangle 307 

Factoring 314 

XIX.  Areas     of     Polygons.     Area     of     the 

Circle.      Proportionality   of   Areas  322 

Areas  of  Polygons 322 

Area  of  the  Circle  . 325 

Proportionality  of  Areas 330 

Problems  of  Construction 335 


STUDY  HELPS  FOR  STUDENTS^ 

The  habits  of  study  formed  in  school  are  of  greater  impor- 
tance than  the  subjects  mastered.  The  following  suggestions, 
if  carefully  followed,  will  help  you  make  your  mind  an  efficient 
tool.  Your  daily  aim  should  be  to  learn  your  lesson  in  less 
time,  or  to  learn  it  better  in  the  same  time. 

1.  Make  out  a  definite  daily  program,  arranging  for  a  definite 
thne  for  the  study  of  mathematics.  You  will  thus  form  the 
habit  of  concentrating  your  thoughts  on  the  subject  at  that 
time. 

2.  Provide  yourself  with  the  material  the  lesson  requires; 
have  on  hand  textbook,  notebook,  ruler,  compass,  special 
paper  needed,  etc.  When  writing,  be  sure  to  have  the 
Ught  from  the  left  side. 

3.  Understand  the  lesson  assignment.  Learn  to  take  notes 
on  the  suggestions  given  by  the  teacher  when  the  lesson  is 
assigned.  Take  dowTi  accurately  the  assignment  and  any 
references  given.  Pick  out  the  important  topics  of  the 
lesson  before  beginning  your  study. 

4.  Learn  to  use  your  textbook,  as  it  will  help  you  to  use  other 
books.  Therefore  understand  the  purpose  of  such  devices 
as  index,  footnotes,  etc.,  and  use  them  freely. 

5.  Do  not  lose  time  getting  ready  for  study.  Sit  down  and 
begin  to  work  at  once.  Concentrate  on  your  work,  i.e.,  put 
your  mind  on  it  and  let  nothing  disturb  you.  Have  the 
will  to  learn. 

1  These  study  helps  are  taken  from  Study  Helps  for  Students 
in  the  University  High  School.  They  have  been  found  to  be  very 
valuable  to  students  in  learning  how  to  study  and  to  teachers  in 
training  students  how  to  study  effectively. 


XX  STUDY  HELP  FOR  STUDENTS 

6.  As  a  rule  it  is  best  to  go  over  the  lesson  quickly,  then  to  go 
over  it  again  carefully;  e.g.,  before  beginning  to  solve  a 
problem  read  it  through  and  be  sure  you  understand  what 
is  given  and  what  is  to  be  proved.  Keep  these  two  things 
clearly  in  mind  while  you  are  working  on  the  problem. 

7.  Do  individual  study.  Learn  to  form  your  own  judgments, 
to  work  your  own  problems.  Individual  study  is  honest 
study. 

8.  Try  to  put  the  facts  you  are  learning  into  practical  use  if 
possible.  Apply  them  to  present-day  conditions.  Illus- 
trate them  in  terms  famihar  to  you. 

9.  Take  an  interest  in  the  subject.  Read  the  corresponding 
hterature  in  your  school  library.  Talk  to  your  parents 
about  your  school  work.  Discuss  with  them  points  that 
interest  you. 

10.  Review  your  lessons  frequently.     If  there  were  points  you 
^  did  not  understand,  the  review  will  help  you  to  master  them. 

11.  Prepare  each  lesson  every  day.    The  habit  of  meeting  each 
requirement  punctually  is  of  extreme  importance. 


CHAPTER  I 

ASSUMPTIONS,  THEOREMS,  AND  CONSTRUCTIONS 
GIVEN  IN  FIRST- YEAR  MATHEMATICS 

To  the  Student 

In  the  first-year  course  the  student  has  become  familiar 
with  a  number  of  geometric  truths.  In  the  second-year 
course  these  are  used  to  estabHsh  other  trtiths.  A  com- 
plete Hst  of  the  geometric  assumptions  and  theorems  of 
the  first  course  is  given  below.  Future  references  will 
be  made  to  this  list,  to  save  the  student  the  inconven- 
ience of  looking  them  up  in  First-Year  Mathematics. 

The  numbers  in  the  parentheses  refer  to  the  sections 
in  First-Year  Mathematics  in  which  the  statements  were 
given  for  the  first  time. 

Classes  which  did  not  use  First-Year  Mathematics  as 
the  text  of  the  first  year  may  use  this  list  as  a  syllabus, 
the  students  working  out  the  proofs  under  the  teacher's 
direction  and  in  the  order  indicated  by  the  numbers  in 
the  parentheses.  For  this  book,  however,  these  truths 
play  the  part  of  assumptions. 

Assumptions 

1.  Through  two  points  one  and  only  one  straight  line 
can  be  drawn.     (20) 

2.  A  straight  line  two  of  whose  points  lie  in  a  plane, 
lies  entirely  in  the  plane.     (204) 

3.  The  shortest  distance  between  two  points  is  the 
straight  line-segment  joining  the  points.     (21) 

4.  Two  straight  lines  intersect  in  one  and  only  one 

point.     (25) 

1 


2  SEC0N1>.YEAII  •^MATHEMATICS 

6.  A  line-segment  or  an  angle  is  equal  to  the  sum  of  all 
its  parts.     (33) 

6.  A  segment  or  an  angle  is  greater  than  any  of  its 
parts,  if  only  positive  magnitudes  are  considered.     (34) 

7.  If  the  same  number  is  added  to  equal  nimibers,  the 
sums  are  equal.     (35) 

8.  If  equals  are  added  to  equals,  the  sums  are 
equal.     (36) 

9.  If  the  same  number  or  equal  numbers  be  subtracted 
from  equal  numbers,  the  differences  are  equal.     (41) 

10.  The  sums  obtained  by  adding  unequals  to  equals 
are  unequal  in  the  same  order  as  are  the  unequal 
addends.     (42) 

11.  The  sums  obtained  by  adding  unequals  to  imequals 
in  the  same  order,  are  unequal  in  the  same  order.     (43) 

12.  The  differences  obtained  by  subtracting  unequals 
from  equals  are  unequal  in  the  order  opposite  to  that  of 
the  subtrahends.     (44) 

13.  If  equals  be  divided  by  equal  numbers  (excluding 
division  by  0),  the  quotients  are  equal.     (78) 

14.  If  equals  be  multiplied  by  the  same  number  or 
equal  numbers,  the  products  are  equal.     (80) 

Angles 

15.  All  right  angles  are  equal.     (118) 

16.  Equal  central  angles  in  the  same  or  equal  circles 
intercept  equal  arcs.     (124) 

17.  In  the  same  or  equal  circles  equal  arcs  are  inter- 
cepted by  equal  central  angles.     (125) 

18.  A  central  angle  is  measured  by  the  intercepted 
arc.     (126) 


ASSUMPTIONS,  THEOREMS,  AND  CONSTRUCTIONS    3 

19.  If  two  angles  have  their  sides  parallel  respectively 
they  are  equal  or  supplementary.     (197) 

20.  If  the  sum  of  two  adjacent  angles  is  a  straight 
angle,  the  exterior  sides  are  in  the  same  straight 
line.     (177) 

21.  The  sum  of  all  the  adjacent  angles  about  a  point, 
on  one  side  of  a  straight  line,  is  a  straight  angle.     (179) 

22.  The  sum  of  all  the  angles  at  a  point  just  covering 
the  angular  space  about  the  point  is  a  perigon.     (180) 

23.  If  two  lines  intersect,   the  opposite  angles  are 

equal.     (183) 

Angles  of  a  Triangle 

24.  The  sum  of  the  angles  of  a  triangle  is  180°. 
(112),  (198) 

25.  The  sum  of  the  exterior  angles  of  a  triangle,  taking 
one  at  each  vertex,  is  360°.     (115) 

26.  An  exterior  angle  of  a  triangle  equals  the  sum  of 
the  two  remote  interior  angles.     (118),  (199) 

27.  If  the  angles  of  one  triangle  are  respect- 
ively equal  to  the  angles  of  another,  the  triangles  are 
similar.     (233) 

28.  The  base  angles  of  an  isosceles  triangle  are 
equal.     (280) 

29.  An  equilateral  triangle  is  equiangular.     (281) 

30.  If  two  angles  of  a  triangle  are  equal  the  triangle 
is  isosceles.     (281) 

31.  The  acute  angles  of  a  right  triangle  are  com- 
plementary angles.     (184) 

32.  In  a  right  triangle  whose  acute  angles  are  30°  and 
60°,  the  side  opposite  the  90°-angle  is  twice  as  lon^  as  the 
side  opposite  the  30°-angle.     (185) 


7 


4  SECOND-YEAR  MATHEMATICS 

33.  If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  to  them  are  unequal,  the  greater  angle  lying 
opposite  the  greater  side.     (281) 

34.  If  two  angles  of  a  triangle  are  unequal  the  sides 
opposite  to  them  are  unequal,  the  greater  side  lying 
opposite  the  greater  angle.     (281) 

Perpendicular  Lines 

35.  The  shortest  distance  from  a  point  to  a  line  is  the 
perpendicular  from  the  point  to  the  line.     (285) 

36.  At  a  given  point  in  a  given  line  one  and  only  one 
perpendicular  can  be  drawn  to  the  line.     (176) 

From  a  given  point  one  and  only  one  perpendicular 
can  be  drawn  to  a  given  line. 

37.  All  points  on  the  perpendicular  bisector  of  a  line- 
segment  are  equidistant  from  the  endpoints  of  the  seg- 
ment.    (281) 

38.  If  a  point  is  equidistant  from  the  endpoints  of  a 
line-segment,  it  is  on  the  perpendicular  bisector  of  the 
segment.     (283) 

39.  If  each  of  two  points  on  one  line  is  equidistant 
from  two  points  of  another  line  the  line's  are  perpendic- 
ular.    (283) 

Parallel  Lines 

40.  Parallel  lines  are  everywhere  equally  distant.  (192) 

41.  One  and  only  one  parallel  can  be  drawn  to  a  line 
from  a  point  outside  the  line.     (194) 

42.  If  two  lines  are  cut  by  a  transversal  making  the 
corresponding  angles  equal,  the  lines  are  parallel.     (195) 

43.  Two  lines  perpendicular  to  the  same  line  are 
parallel.     (195) 

44.  Two  lines  are  parallel  if  two  alternate  interior 
angles  formed  with  a  transversal  are  equal.     (195) 


ASSUMPTIONS,  THEOREMS,  AND  CONSTRUCTIONS    5 

45.  Two  lines  are  parallel  if  the  interior  angles  on 
the  same  side  formed  with  a  transversal  are  supple- 
mentary.    (195) 

46.  Two  lines  parallel  to  the  same  line  are  parallel 
to  each  other.     (195) 

47.  If  two  parallel  lines  are  cut  by  a  transversal  the 
corresponding  angles  are  equal;  the  alternate  interior 
angles  are  equal;  the  interior  angles  on  the  same  side  are 
supplementary.     (196) 

Proportional  Line-Segments 

48.  A  line  parallel  to  one  side  of  a  triangle  divides 
the  other  two  sides  into  corresponding  parts  having  equal 
ratios.     (244)  / 

49.  A  line  bisecting  an  angle  of  a  triangle  divides  the 
side  opposite  that  angle  into  parts  whose  ratio  is  equal 
to  the  ratio  of  the  other  sides.     (245) 

50.  A  line  dividing  two  sid^s  of  a  triangle  into  cor- 
responding parts  having  the  same  ratio,  is  parallel  to  the 
third  side  of  the  triangle.     (246) 

Areas  and  Volumes  • 

51.  The  area  of  a  square  is  equal  to  the  square  of  one 
side.     (140) 

52.  The  area  of  a  rectangle  equals  the  product  of  the 
base  by  the  altitude.     (141) 

53.  The  volume  of  a  rectangular  parallelopiped  equals 
the  product  of  the  length  by  the  height  by  the  width.    (145) 

54.  The  volume  of  a  cube  is  equal  to  the  cube  of  one 
edge.     (146) 

55.  The  area  of  a  parallelogram  is  equal  to  the  pro- 
duct of  the  base  by  the  altitude.     (163) 


6  SECOND-YEAR  MATHEMATICS 

56.  The  area  of  a  triangle  is  equal  to  one-half  the 
product  of  the  base  by  the  altitude.     (164) 

57.  The  area  of  a  trapezoid  is  equal  to  one-half  the 
■product  of  the  altitude  by  the  sum  of  the  bases.     (166) 

Proportionality  of  Areas 

58.  In  a  proportion  the  product  of  the  means  is  equal 
to  the  product  of  the  extremes.     (259) 

59.  The  areas  of  two  rectangles  are  in  the  same  ratio 
as  the  products  of  their  dimensions.     (260) 

60.  Two  rectangles  having  equal  bases  are  in  the  same 
ratio  as  the  altitudes.     (261) 

61.  Two  rectangles  having  equal  altitudes  are  in  the 
same  ratio  as  the  bases.     (262) 

62.  The  areas  of  parallelograms  are  in  the  same  ratio 
as  the  products  of  the  bases  and  altitudes.     (263) 

63.  The  areas  of  triangles  are  in  the  same  ratio  as  the 
products  of  the  bases  and  altitudes.     (264) 

64.  The  areas  of  parallelograms  having  equal  bases 
are  in  the  same  ratio  as  the  altitudes.     (265) 

65.  The  areas  of  triangles  having  equal  bases  are  in 
the  same  ratio  as  the  altitudes.     (266) 

Congruent  Triangles 

66.  Two  triangles  are  congruent  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other,     (s.a.s.)     (274) 

67.  Two  triangles  are  congruent  if  two  angles  and  the 
side  included  between  their  vertices  in  one  triangle  are 
equal  respectively  to  the  corresponding  parts  in  the  other. 
(a.s.a.)  (275) 

68.  If  three  sides  of  one  triangle  are  equal,  respec- 
tively, to  the  three  sides  of  another  triangle,  the  triangles 
are  congruent,     (s.s.s.)     (283) 


ASSUMPTIONS,  THEOREMS,  AND  CONSTRUCTIONS    7 

69.  Two  right  triangles  are  congruent  if  the  hypote- 
nuse and  one  side  of  one  are  equal  respectively  to  the 
hypotenuse  and  a  side  of  the  other.     (285) 

Similar  Triangles 

70.  Two  triangles  are  similar  if  the  ratios  of  the  cor- 
responding sides  are  equal.     (236) 

Loci 

71.  The  perpendicular  bisector  of  a  segment  is  the 
locus  of  all  points  equidistant  from  its  endpoints.     (284) 

72.  The  bisector  of  an  angle  is  the  locus  of  points 
which  are  equidistant  from  the  sides.     (304) 

Tangents 

73.  The  radius  drawn  to  the  point  of  contact  of  a 
tangent  is  perpendicular  to  the  tangent.     (308) 

74.  A  line  perpendicular  to  a  radius  at  the  outer 
endpoint  is  tangent  to  the  circle.     (309) 

Theorem  of  Pythagoras 

75.  In  a  right  triangle  the  sum  of  the  squares  on  the 
sides  including  the  right  angle  is  equal  to  the  square 
on  the  hypotenuse.     (402) 


CHAPTER  II 

METHODS  OF  PROOF 

Logic 

76.  Reasoning.  In  the  first-year  course  we  studied 
some  of  the  laws  of  algebra  and  became  acquainted  with 
a  number  of  useful  geometric  facts.  The  truth  of  many 
of  these  facts  was  found  and  verified  by  measurement,  of 
others,  especially  toward  the  end  of  the  course,  by  a 
process  of  reasoning,  or  proof. 

In  our  everyday  life  we  reason  whenever  we  infer  one 
truth  from  another.  Thus,  from  the  general  truths  that 
metals  are  good  conductors  of  heat  and  that  aluminium 
is  a  metal,  we  infer  that  aluminium  is  a  good  conductor 
of  heat.  Or,  if  we  accept  as  true  the  statement  that  iron 
is  the  most  useful  metal  and  that  iron  is  the  cheapest 
metal,  we  may  infer  that  the  most  useful  metal  is  also 
the  cheapest  metal. 

In  every  branch  of  knowledge  there  are  employed 
certain  principles  and  forms  of  thought  by  means  of  which 
all  persons  must  think  and  reason.  Logic  treats  of  these 
principles.  Moreover,  it  helps  us  to  avoid  the  fallacies 
which  may  arise  from  neglecting  the  correct  rules  of 
thinking.  In  particular,  it  p^in+s  out  why  it  is  absurd 
to  make  such  an  inference  as  that  all  Europeans  are 
Frenchmen  from  the  known  fact  that  all  Frenchmen  are 
Europeans. 

False  reasoning.  By  incorrect  reasoning,  some  of  the 
ancient   Greek   philosophers   pretended   to   prove   that 

8 


METHODS  OF  PROOF  9 

motion  was  impossible.  "For,"  they  said,  "a,  moving 
body  must  move  either  in  the  place  where  it  is,  or  where 
it  is  not;  now  it  is  absurd  to  hold  that  a  body  could  be 
where  it  is  not;  and  if  it  moves,  it  cannot  be  in  a  place 
where  it  is;  therefore  it  cannot  move  at  all." 

The  student  is  probably  familiar  with  the  following 
absurdity : 

No  dog  has  9  tails.         ^  ^ 

One  dog  has  1  more  tail  than  no  dog. 

Therefore,  one  dog  has  10  tails. 

Thus,  to  know  how  to  use  the  rules  of  correct  reason- 
ing is  valuable  also  in  that  it  enables  us  to  point  out  weak 
places  in  an  incorrect  argument,  and  to  replace  incorrect 
reasoning  by  sound  reasoning  in  our  own  work. 


Geometrical  Fallacies 

77.  Likewise  the  reasoning  used  in  geometry  and 
algebra  follows  certain  laws.  The  importance  of  exer-' 
cising  great  care  in  a  geometric  proof  may  be  illustrated 
by  two  of  the  well-known  puzzles  of  geometry,  viz. : 

1.  Theorem:  Every  triangle  is  isosceles. 

Given  any  triangle,  as 
ABC,  Fig.  1. 

To  prove  that  ABC  is 
isosceles. 

Proof:  Let  D^  be  the 
perpendicular  bisector  oi  AB 
and  let  CE  be  the  bisector^of 
angle  C,  meeting  DE  at  E. 

From  E  draw  EA   and  EB 


10 


SECOND-YEAR  MATHEMATICS 


Draw    EG  perpendicular  to  AC,  and  EF  perpen- 
dicular to  CB. 

Then  A  ADE  ^  A  BDE  (§  69). 

Hence,  AE=BE 

(since  corresponding  sides  of  congruent  triangles  are  equal). 

ACEG^ACEF  (^Q7). 
Hence,  EG = EF  and  CG  =  CF.    Why  ? 

Therefore  A  AEG^  A  BEF 

(hypotenuse  and  a  side,  §  69). 

Hence,  GA=FB. 

Since  CG  =  CF, 

it  follows  that  CG+GA=CF+FB, 
or     CA==^CB. 

Therefore  the  triangle  ABC, 
although  known  not  to  be  isos- 
celes, would  seem  to  have  been 
proved  to  be  isosceles. 

Make  a  careful  construction  of  Fig.  1,  and  discover 
'  the  error  in  the  demonstration. 

2.  To  show  geometrically  that  64  =  65. 

Draw  two  right  triangles  having  the  sides  including 
the  right  angle  equal  to  3  and  8,  respectively  (Fig.  2). 


Fig.  2 


Fig.  3 


Fig.  4 


Draw   two  quadrilaterals    (Fig.  3)   having  one  pair  of 
opposite  sides  parallel  and  equal  to  3  and  5,  respectively, 


METHODS  OF  PROOF 


11 


and  the  third  side  perpendicular  to  the  parallel  sides 

and    equal    to    5.     Placing   the 

triangles   and    quadrilaterals   as 

in  Fig.  4,  a  square  is  obtained 

whose  area  is  equal  to  8X8=64. 

If  now,  they  are  placed  as  in 

Fig.   5,   a   rectangle    is    formed 

whose  area  is  equal  to  13X5  =65. 

Hence,  64=65! 


II  1^     I 2.J__ 


Fig.  5 


Make  a  careful  construction  and  discover  the  error. 

78.  Need  for  proof.  In  both  the  fallacies  in  §  77  the 
difficulty  has  come  from  assuming  that  what  looks  to 
be  nearly  true  is  exactly  true.  The  moral  is,  of  course, 
things  that  look  correct  cannot  always  be  relied  upon  as 
correct.  The  word  'intuition''  is  used  to  designate  the 
sort  of  reasoning  that  draws  its  conclusions  from  direct 
appearances. 

The  following  exercises  are  illustrations  of  the  danger  of 
going  astray  even  in  geometry  through  too  ready  a  reliance 
on  intuition. 

EXERCISES 

1.  Compare  the  segments  a  and  b,  Fig.  6,  as  to  length  by 
looking  at  the  figure.     Then  measure 
each  segment. 


> 


<- 


-> 


Fig.  6 


b 

Fig.  7 


2.  Compare,  as  in  Exercise  1,  the  segments  a  and  b  in 
Fig.  7.    Test  by  measuring. 


12 


SECOND-YEAR  MATHEMATICS 


3.  Are  the  lines  AB  and  CD  in 
Fig.  8  parallel  ?  Answer  the  question, 
then  test  by  measuring  the  distances 
between  the  hues. 


^ ///////////////// B 

C\\\\\\\\\\\\\\D 
Fig.  8 


4.  Are  the  lines  AB  and  CD,  Figs.  9  and  10,  in  the  same 
straight  line?    Test  with  a  ruler. 


B        c 
Fig.  9 


D 


5.  Are   the   lines   AB   and  CD,   Fig.    11,   straight  Hnes? 
Test  with  a  ruler. 


Fig.  11 

6.  Count  the  number  of  blocks 
in  Fig.  12.  Continue  to  look  at  the 
figure  and  you  will  see  either  one 
more  or  one  less. 


Fig.  12 


79.  Methods.  There  is  no  one  specific  method  by 
which  all  theorems  or  problems  may  be  attacked  or 
proved.  However,  certain  general  directions  and  methods 
as  to  the  way  of  attacking  problems  and  proving  theorems 
may  be  stated.  A  knowledge  of  these  methods  is  of 
greatest  importance  as  they  will  keep  the  student  from 
groping  about  blindly  for  a  proof,  wasting  his  time  and 
energy.  Several  methods  of  proof  are  discussed  in  this 
chapter,  others  are  considered  in  chapter  IV. 


METHODS  OF  PROOF  13 

80.  General  directions.    Hypothesis.    Conclusion. 

1.  Read  the  problem  carefully,  get  it  clearly  in  mind, 
and  keep  it  in  mind  while  at  work  on  it.  Most  problems 
need  at  least  two  readings. 

2.  If  the  problem  is  a  geometric  theorem  or  exercise, 
draw  carefully  a  general  figure.  Thus,  if  the  theorem  refers 
to  a  triangle,  draw  a  triangle  with  unequal  sides,  not  an 
equilateral,  or  isosceles,  or  right  triangle.  This  will  keep 
you  from  conunitting  the  error  of  proving  a  theorem  only 
for  a  special  case. 

3.  Write  down  what  is  given  (the  hypothesis)  and  what 
is  to  be  proved  (the  conclusion),  referring  all  statements 
to  the  figure. 

4.  If  a  proof  does  not  readily  suggest  itself  to  you, 
think  of  all  the  things  you  have  learned  that  are  like  the 
problem  you  are  trying  to  work  out,  e.g.,  recall  the 
theorems  that  seem  like  the  task  before  you. 

Thus,  if  you  are  to  prove  two  angles  equal,  ask  the 
question:  Under  what  conditions  are  two  angles  equal? 
If  you  wish  to  prove  two  lines  parallel,  the  question  should 
be :  When  are  two  lines  parallel  ?  Then  select  the  theorem 
that  seems  to  you  most  promising  or  suitable,  until  you 
find  something  that  brings  you  to  your  goal.  It  is  a  good 
plan  to  review  and  summarize  the  theorems  and  prob- 
lems that  have  been  established  previously.  Keep  up 
this  practice  until  it  becomes  a  hahit,  and  you  will  acquire 
the  art  of  selecting  very  quickly  the 
theorem  that  is  needed  to  prove  a 
new  theorem  or  problem.  , 

5.  The  conclusion  may  sometimes  / 
be  obtained  by  drawing  lines,  not  given          / 
in  the  figure,  as  described  by  the  hy- 
pothesis.    Thus,  if  AC==CB,  Fig.  13,  Fig.  13 


14 


SECOND-YEAR  MATHEMATICS 


we  may  prove  that  Z  A  =  ZB,  hy  drawing  the  bisector  of 
angle  C  and  then  proving  A  ADC^  A  BDC. 

81.  Method  of  proof  by  superposition.  This  method 
was  used  in  proving  some  of  the  theorems  on  congruent 
triangles,  §§  66,  67.  It  consists  in  placing  one  figure 
over  another  and  then  showing  that  all  parts  of  the  one 
coincide  with  the  corresponding  parts  of  the  other. 
This  method,  although  practical  when  tjie  elements 
involved  in  the  proof  are  few  and  specific,  is  not  considered 
a  good  theoretical  test  by  the  mathematician.  For,  the 
axioms  validating  superposition  are  usually  not  given 
in  full  detail.  The  result  is  that  the  student  is  in  danger 
of  drawing  rashly  the  conclusion  which  is  to  be  estabhshed 
by  the  superposition  of  the  one  figure  upon  the  other. 
The  method  is  used  only  in  a  few  cases. 


82.  Method  of  congruent  triangles. 

When  trying  to  prove  that  lines  or 
angles  are  equal,  it  is  sometimes 
possible  to  show  that  they  are  corre- 
sponding parts  of  congruent  triangles. 
It  may  be  necessary  to  draw  help- 
ing lines  to  obtain  the  congruent 
triangles,  of  which  the  lines  or  angles 
to  be  proved  equal  are  corresponding 
parts.  The  following  proof  will  illus- 
trate the  method: 


^^ 


^D 


y 


Fig.  14 


83.  Theorem:  If  each  of  two  points  on  a  given  line  is 
equally  distant  from  two  given  points,  the  given  line  is  the 
perpendicular  bisector  of  the  segment  joining  the  given  points. 

Given  the  line,  AB,  Fig.  14,  and  the  points  C  and  D 
such  that 

AC=AD,  CB=BD 

To  prove  x=x\      CE=ED 


METHODS  OF  PROOF  15 

Preliminary   discussion:     We   know   that    x=x' ^    if 

However,  since  we  only  know  that  AE=AE,  that  CA  =AD 
and  therefore  that  ZACE=^ADE  (§28),  we  do  not 
have  the  required  parts  to  show  that  A  CAE^  A  DAE. 
Hence,  we  shall  first  prove  y  =  y',  by  proving  that 
AACB^AADB. 

Proof: 

STATEMENTS  REASONS 

AC=AD,  CB= ED... by  hypothesis. 

AB  =  AB common  to  both  triangles 

ACB  and  ADB. 

Therefore,  A  ACB  ^  A  ADB s.s.s.  (68) 

Hence,  y  =  y' corresponding    parts     of 

congruent    triangles    are 
equal. 

AE  =  AE common. 

AC=AD by  h}T)othesis. 

Therefore,  AACE^  A  ADE s.a.s.  (67) 

Hence,  x  =  z\  and  CE = ED. .  corresponding    parts    of 

congruent    triangles    are 
equal. 

84.  Sjmibols  for  "therefore"  and  "since."  The  symbol 
/.  means  therefore,  and  *.•  means  since. 

85.  Conventional    treatment   of    a    theorem.      The 

formal  demonstration  of  a  theorem  consists  of  three 
main  parts:  the  hypothesis,  conclusion,  and  proof.  In 
writing  a  proof  a  reason  must  be  given  for  each 
step.  This  means  that  each  statement  must  be  based 
upon  (1)  a  definition,  (2)  the  hypothesis,  (3)  an 
axiom,    or    (4)    a    theorem    which    has    been    proved 


16  SECOND-YEAR  MATHEMATICS 

previously.*    The  last  step  in  the  proof  must  be  the 
same  as  the  conclusion,  f 

86.  Reviews.  It  is  a  good  plan  to  review  daily  for  a 
time  after  passing  them,  the  proofs  of  theorems  previ- 
ously established.  This  may  be  done  by  simply  recalling 
the  figure,  the  method  of  proof  used,  and  the  principal 
steps,  i.e.,  a  sort  of  sketch  or  outline  of  the  proof.  Thus, 
in  a  few  minutes  a  day  the  student  will  accomplish  easily 
what  will  be  a  most  difficult  task  if  left  until  the 
end  of  a  chapter,  or  until  the  day  before  an  examination. 

♦Hippocrates  (b.  about  470 B.C.)  introduced  the  method  of 
"reducing"  one  theorem  to  another  that  has  been  previously 
proved.  See  W.  W.  R.  Ball,  A  Short  Account  of  the  History  of  Mathe- 
matics, 5th  ed.,  p.  39,  hereafter  referred  to  as  Ball. 

t  The  processes  of  proving  theorems  were  developed  by  the 
Greeks.  Greece  was  indebted  to  Egypt  for  its  beginnings  in  geome- 
try. However,  the  Egyptians  carried  geometry  no  farther  than  was 
necessary  for  the  practical  needs  of  life.  They  may  have  felt  the 
truth  of  some  theorems;  but  the  Greeks  formulated  these  geometric 
truths  into  scientific  language  and  subjected  them  to  proof  (see 
Ball,  pp.  16-19).  The  Greeks  also  recognized  that  it  is  impossible 
to  prove  everything  in  geometry  and  that  some  simple  statements  have 
to  be  assumed. 

Euchd  (about  300  B.C.)  used  the  term  common  notion  in  the 
sense  in  which  in  modern  mathematics  we  use  axiom,  i.e.,  a  general 
statement  admitted  to  be  true  without  proof.  Thus,  the  statement : 
"If  equals  are  added  to  equals,  the  sums  are  equal"  is  an  axiom 
because  it  holds  in  mathematics  in  general,  i.e.,  in  arithmetic, 
algebra,  or  geometry. 

In  modern  mathematics,  a  statement  referring  to  geometry 
only  and  admitted  to  be  true  without  proof,  is  called  a  postulate. 
Thus,  the  statement  "Two  points  determine  a  straight  line"  is  a 
postulate.  Some  textbook  writers  use  the  word  axiom  or  assuynp- 
iion  to  denote  postulates  as  well  as  axioms. 

Moreover,  just  as  we  assume  unproved  propositions,  we  have 
undefined  terms,  such  as  points,  lines,  etc.  Pasch  (1881)  recognized 
the  obvious  impossibility  of  defining  everything  in  geometry. 


METHODS  OF  PROOF 


17 


87.  Inductive  method.  Mathematical  facts  can  often 
be  discovered  by  considering  enough  special  cases  to  enable 
the  student  to  recognize  the  general  law  underlying  these 
cases.  The  method  may  be  illustrated  by  the  following 
example : 

EXAMPLE   OF   INDUCTIVE   METHOD 

Problem:  It  is  known  that  the  sum  of  the  angles  of 
a  triangle  is  180°.  What  is  the  sum  of  the  angles  of  a 
quadrilateral,  pentagon  ....  ,  etc.,  or  of  any  polygon? 

To  find  the  sum  of  the  angles  of  a  polygon,  divide 
it  into  triangles  by  means  of  diagonals  drawn  from  one 


Fig.  15 


Fig.  16 


Fig.  17 


vertex  to  the  others.  Thus,  a  quadrilateral  may  be 
divided  into  two  triangles,  Fig.  15;  a  pentagon  into  three 
triangles,  Fig.  16;  a  hexagon  into  four  triangles.  Fig.  17, 
etc.  The  table  below  gives  the  sum  of  the  angles  in 
the  various  cases. 

How  does  the  number  of  triangles  in  each  polygon 
compare  with  the  number  of  sides?  Hence,  how  does 
the  sum  of  the  angles  compare  with  the  number  of  sides  ? 

What  seems  to  be  the  sum  of  the  angles  of  an  n-gon  ? 

Make  the  table  complete  by  filling  out  the  blank  spaces. 


Number  of  sides  of 
polygon 

3 

4 

5 

6 

7 

10 

15 

n 

Number  of  triangles 

1 

2  ^ 

3 

4 

5 

Sum  of  angles 

180° 

2X180° 

3X180° 

4X180° 

5X180° 

18 


SECOND-YEAE  MATHEMATICS 


It  is  seen  that  the  inductive  method  suggests  mathematical 
facts,  but  does  not  prove  them.  Hence,  having  found  that 
the  sum  of  the  interior  angles  of  an  n-gon  would  seem  to 
be  (n— 2)  180°,  it  still  remains  to  be  proved  that  this  is 
true.    This  may  be  done  as  follows: 


Fig.   18,  having 


88.  Theorem:  The  sum  of  the  interior  angles  of  a 
polygon  having  n  sides  is  {n— 2)180°,  or  {n—2)  straight 
angles. 

Given  the  polygon  ABCD  .  .  . 
n  sides. 

To  prove  that  the  sum  of  the 
interior  angles,  S,  is  given  by  the 
equation: 

>S  =  (n-2)180°. 

Proof :  Draw  diagonals  from  A 
to  the  other  vertices. 

This  divides  the  polygon  into  Fig.  18 

(n—2)    triangles.     Why? 

The  sum  of  the  angles  of  the  triangles  of  the  polygon 
is  (n -2)  180°.     Why? 

The  sum  of  the  angles  of  these  triangles  is  equal  to 
the  sum  of  the  angles  of  the  polygon.     Why  ? 

Hence,  the  sum  of  the  angles  of  the  polygon  is 
(n-2)180°.     Why? 


EXERCISES 

1.  Using  the  formula  /S=(w— 2)180°,  find  the  sum  of  the 
interior  angles  of  hexagon,  octagon,  decagon,  2n-gon. 

2.  The  sum  of  the  angles  of  a  polygon  is  1800°.     Find  the 
number  of  sides. 


METHODS  OF  PROOF 


19 


89.  Theorem:  The  sum  of  the  exterior  angles  of  O/ 
polygon,  one  exterior  angle  at  each  vertex  being  taken,  is 
360°,  or  2  straight  angles.  l 

Given  the  polygon  ABCD  .  .  .  .  ,  etc.,  Fig.  19, 
having  n  sides  and  the  exterior  angles  a,  b,  c,  d  .  .  .  .  , 
etc. 

To  prove  that  a+^+cH-^  .  .  .  .  =  360°,  or  2  straight 
angles. 


Preliminary  discussion : 

How  is  an  exterior  angle 
related  to  the  adjacent  interior 
angle  ? 

How  may  we  j&nd  the  sum 
of  the  exterior  and  interior 
angles  ? 

Knowmg  the  sum  of  the  in- 
terior angles  to  be  (n— 2)180°, 
how  may  we  find  the  sum  of 
the  exterior  angles? 


^ 


H 


g' 


f/f 


K 


h'/h 


Fig.  19 


Proof: 


a+a'  =  180°  Why? 

6+y  =  180°  Why? 

c+c'  =  180°,  Why? 

etc. 


a^b^c-V.-ol-^b'-Vc'-^. .  =71-180°        =(180n)^ 
Why? 

a!  ^y  ■\-c' -\- . .  =(7i-2)180°  =  (180n)°-360' 
Why? 


a-\-b-\-c. . 


=  360°     Why? 


Show  that  the  sum  of  the  exterior  angles  of  a  polygon 
is  independent  of  the  number  of  sides  of  the  polygon. 


20  SECOND-YEAR  MATHEMATICS 


EXERCISES 

1.  Prove  that  any  interior  angle  of  a  regular  polygon  is 

2.  If  two  angles  of  a  quadrilateral  are  supplementary,  show 
that  the  other  two  are  supplementary. 

3.  How  many  right  angles  are  contained  in  the  sum  of  the 
angles  of  a  polygon  having  n  sides  ? 

4.  How  many  sides  has  a  polygon  the  sum  of  whose  angles 
is  36  right  angles  ?     18  straight  angles  ?    720°  ? 

5.  Show  that  the  sum,  S,  of  the  interior  angles  of  a  polygon 
is  &fu7iction  of  the  number  of  sides. 

6.  What  is  the  sum  of  the  vertex  angles  a,  6,  c,  d,  and  e  of  the 
five-point  star,  Fig.  20  ? 

7.  If  the  sum  of  the  interior  angles  of 
a  polygon  is  twice  the  sum  of  the  exterior 
angles,  how  many  sides  are  there  in  the 
polygon  ? 

8.  How  many  diagonals  may  be  drawn 
in  a  polygon  having  4  sides?  5  sides? 
6  sides?  Fi«-20 

9.  Show  that  in  an  w-gon  (n— 3)  diagonals  may  be  drawn 
from  one  vertex. 

10.  Show  that  in  an  n-gon  — ^ —  diagonals  may  be  drawn. 

11.  Show  that  the  number  of  diagonals,  N,  that  may  be 
drawn  in  a  polygon  is  a  function  of  the  number  of  sides,  n. 

90.  Algebraic  method.  This  method  is  used  when 
the  numerical  value  of  a  magnitude  is  to  be  found  or 
when  a  relation  between  several  magnitudes  is  to  be 
proved. 

First,  the  relations  between  the  magnitudes  are 
expressed  in  algebraic  symbols.     The  required  magnitude 


METHODS  OF  PROOF  21 

is  then  found  by  a  process  of  elimination.     The  following 
problem  illustrates  the  method. 

EXAMPLE    OF   ALGEBRAIC    METHOD 

Prove  that  the  bisectors  of  two  supplementary  ad- 
jacent angles  are  perpen- 
dicular to  each  other. 

Given   that  x  and  y,                  j 
Fig.  21 ,  are  adj  acent  angles               / 
and  that  x-\-y  =  180.  .' 

Also,  a  =  a'  and  h  =  h'.  Fig.  21 

To  prove  that  a'+6'=90. 

Proof:  a+a'+b' +h  =  lSO.        Why? 

Thus,  we  have  a  relation  between  a,  a',  h  and  b'. 

Since  the  conclusion  contains  only  a'  and  6',  we  must 
eliminate  a  and  b  from  the  equation  a-f  a'+6'-f-6  =  180. 

a  =  a'        Why? 
and  b  =  b'        Why? 

Then,  a  and  b  may  be  eliminated  by  substituting  a'  and  6' 
for  a  and  6,  respectively. 

This  gives  a'+a'+&'+?>'  =  180 
Collecting  terms,  2a'  +26'  =  180. 
Hence,  a'+¥=   90      Why? 

91.  In  the  preceding  proof  a  and  b  were  eliminated  by 
substitution.  Methods  of  elimination  will  be  discussed 
in  the  next  chapter. 

EXERCISES 

Prove  the  following  exercises : 

1.  If  two  angles  of  one  triangle  are  equal  to  two  angles  of 
another,  the  third  angles  are  equal  and  the  triangles  are  mutually 
equiangular. 


22  SECOND-YEAR  MATHEMATICS 

2.  Find  the  angle  formed  by  the  bisectors  of  the  acute  angles 
of  a  right  triangle. 

3.  One  base  angle  of  an  isosceles  triangle  is  J  of  the  vertex 
angle.     Find  the  angles  of  the  triangle. 


Summary 

92.  The  chapter  has  shown  the  value  of  logic  in 
supporting  correct  reasoning  and  detecting  fallacies,  the 
danger  in  depending  upon  intuition  alone  as  a  means  of 
proof,  and  the  need  for  a  logical  proof. 

93.  The  meaning  of  the  following  terms  has  been 
taught:  hypothesis,  conclusion,  proof. 

94.  The  following  methods  of  proof  have  been  illus- 
trated: superposition,  the  method  of  congruent  triangles, 
the  inductive  method,  and  the  algebraic  method. 

95.  Some  general  directions  have  been  given  for 
attacking,  or  proving,  problems  and  theorems  (§80). 
The  importance  of  systematic  reviews  has  been  empha- 
sized (§86).  In  the  study  helps  (p.  xix)  the  student  will 
find  some  valuable  suggestions  as  to  the  way  he  may  study 
effectively. 

96.  The  following  theorems  have  been  proved: 

1.  The  sum  of  the  interior  angles  of  a  polygon  is  in— 2) 
straight  angles. 

2.  The  sum  of  the  exterior  angles  of  a  polygon  is  360°. 

3.  If  each  of  two  points  on  a  given  line  is  equally  distant 
from  two  given  points,  the  given  line  is  the  perpendicular 
bisector  of  the  segment  joining  the  given  points. 


CHAPTER  III 

METHODS  OF  ELIMINATION.    PROBLEMS  AND 
EXERCISES  IN  TWO  UNKNOWN  NUMBERS 

97.  Elimination.  In  the  first-year  course  we  learned 
how  to  eliminate  literal  numbers  by  addition  or  sub- 
traction. In  §  90  we  have  seen  that  in  a  system  of  equa- 
tions magnitudes  may  be  eliminated  by  substituting 
equal  magnitudes  for  them.  In  future  work  we  shall 
have  occasion  frequently  to  eliminate  numbers.  There 
are  various  methods  of  elimination,  and  we  should  be 
able  to  select  that  method  which  for  any  particular 
problem  is  most  advantageous.  We  shall  accordingly 
review  briefly  what  we  know  about  elimination,  and  then 
study  other  methods. 

98.  Elimination    by    addition    or    subtraction.     The 

solution  of  the  following  system  (pair)  of  equations  will 
recall  the  method  of  eliminating  by  addition  or  sub- 
traction. 

ILLUSTRATIVE   PROBLEM 

Let  9x-  Sy  =  l 

and  15x+12y  =  S 

The  problem  is  to  find  the  values  of  x  and  y. 
Multiplying  the  first  equation  by  3  and  the  second  by 
2,  we  have 

27x-2Ay=  3 

30a;+242/  =  16 

23 


24  SECOND-YEAR  MATHEMATICS 

By  adding  the  equations  the  y-terms  are  ehminated, 
and  we  obtain 

57a;  =  19 

Substituting  this  value  for  x  in  either  one  of  the  given 
equations,  as  9x — 82/  =  1,  we  get 

3-8|/  =  l 
Sy  =  2 

y=l 

.   1 


and  v=-\  ^^^^^  solution  of  the  system. 

Thus,  to  ehminate  by  addition  or  subtraction  we 
proceed  as  follows: 

1.  By  multiplying  one  or  both  equations  by  the  proper 
numbers  the  coefficients  of  one  of  the  unknown  numbers  are 
made  numerically  the  same  in  both  equations. 

2.  07ie  of  the  unknowns  is  then  eliminated  by  adding  or 
subtracting  the  equations  according  as  the  coefficients  of  this 
unknown  have  unlike  signs,  or  like  signs. 


EXERCISES 

Solve  the  following  systems,  eliminating  by  addition  or 
subtraction: 

f27a;-5i/  =  26  j2.7a+S.5b  =  2A 

'  \l8x+77/=131  ^'  \2.7a-3.56=    .3 

f7m+5n  =  81  (^^+^2/=  } 

•  \9m-2n  =  62  *'  \|x+|2/  =  3j- 


METHODS  OF  ELIMINATION 


25 


99.  Graphical  method  of  solving  a  system  of  equations. 
The  pupil  will  recall  that  every  hnear  equation  in  two 


variables,  as  x  and  y, 
by  a  straight  line.  To 
graph  a  linear  equa- 
tion in  two  variables, 
we  may  graph  two, 
preferably  three,  solu- 
tions of  the  equation 
and  draw  the  straight 
line  passing  through 
the  three  points  cor- 
responding to  the  so- 
lutions. 

The  solution  of  a 
system  of  two  linear 
equations  consists  of 
the  X-  and  ^/-distances 
(co-ordinates)   of  the 


may  be  represented  graphically 


Fig.  22 


point  of  intersection  of  the  two  straight  lines.  In  Fig.  22, 
Jine  AB  represents  the  equation  9x—8y  =  l  and  line  CD 
represents  15x4-12?/ =8.  The  point  of  intersection,  P, 
represents  the  solution  x  =  ^,  y  =  j,  in  the  sense  that  the 
X-  and  ^/-distances  of  P  represent  the  values  of  x  and  y 
that  satisfy  9x—Sy  =  l  and  15a; +122/ =  8,  simultaneously. 


EXERCISES 


Solve  graphically  the  following  systems: 

jSx-Ay=U  (x-2y+4.  =  0 

^'  \5x-\-2y  =  S2  +'^-  \x-{-y  =  5 

„    f9x-\-ey  =  51 


t4. 


jSx-2y  =  9 

\Ax+Sy  =  24:  +*•  \2x-3y  =  4: 

t  AH  problems   marked    %  are   not   essential,   and   may   be 
omitted  at  the  discretion  of  the  teacher. 


26  SECOND-YEAR  MATHEMATICS 

100.  Elimination  by  substitution.  This  method  is 
most  advantageous  when  one  of  the  unknown  numbers 
is  easily  expressed  in  terms  of  the  other. 

For  example,  if  x—2y  =  7,  it  follows  that  x  =  7+2y. 
Why? 

The  following  problem  will  illustrate  the  metho(3 : 

ILLUSTRATIVE  PROBLEM 

Solve  the  following  system  of  equations  eliminating  by 
substitution: 

7w-2z  =  4lQ (1) 

w-^z  =  lS (2) 

Solving  equation  (2)  for  z,  z=lS—w 
Substituting  13— w;  for  z  in  equation  (1), 

7w-2{13-w)  =46.    This  eliminates  z. 
Hence,  w  =  8.    Why? 

and  2  =  5.    Why? 

'w  =  8 


The  solution  is      >        - 
z  =  5 

Thus,  to  solve  a  system  of  equations,  eliminating  by 
substitution,  express  one  of  the  unknown  numbers  in  terms 
of  the  other  by  solving  one  equation.  Then  substitute  the 
result  in  the  other  equation,  and  solve  the  equation  thus 
obtained* 

PROBLEMS   AND   EXERCISES 

Solve  the  equations  obtained  from  the  following  problems 
by  the  method  of  elimination  by  substitution :  • 

1.  One  of  the  base  angles,  x,  of  an  isosceles  triangle  is  equal 
to  twice  the  vertex  angle,  y.    Find  all  the  angles  of  the  triangle. 

2.  The  difference  of  two  numbers  is  14,  and  the  sum  is  100. 
What  are  the  numbers  ? 

*This  method  of  solving  equations  was  first  used  by  Isaac 
Newton  (1642-1727). 


METHODS  OF  ELIMINATION  27 

3.  A  man  invests  one  part  of  $3,200  at  6  per  cent  and  the 
other  part  at  5  per  cent.  If  his  annual  income  is  $180,  how  much 
did  he  invest  at  each  rate  ? 

Solve  the  following  systems  of  equations:* 

(9R-2r  =  U  (9x  =  2y-^84: 

*•  [QR-  r=31  •'•      \7x+-y=7S 

.^    /  x+2y=17  .      t7x  =  QS-^y 


3a;-  y  =  2  *      (  x  =  5y-S0 


g    (Sx+oy  =  U  g    /2x=  y+ 


\2x-  y=  2  ^'  \Qx+6y  =  5 

10.  The  angles  r  and  3s  are  supplementary  and  r—s  =  20°. 
Find  r,  s,  and  3s. 

11.  The  angles  x  and  y  are  complementary  and  the  differ- 
ence is  10°.    Find  x  and  y. 

12.  The  sides  of  an  equiangular  triangle  are  denoted  by 
x+Sy,  2x—y,  and  14.     Find  x  and  y. 

13.  The  angles  of  an  equiangular  triangle  are  denoted  by 
7x+2y,  3(3x-22j),  and  60.     Find  x  and  y. 

1 14.  The  three  sides  of  an  equiangular  triangle  are  denoted 
by  7x+2y,  5{x+2y),  and  8x-Sy-\-2.    Fmd  x  and  y. 

15.  A  man  bought  two  pieces  of  vacant  property,  one  at 
$42  a  foot,  the  other  at  $56  a  foot.  Altogether  he  had  140  ft., 
and  paid  for  it  the  sum  of  $6,780.  Find  the  number  of  feet  of 
ground  he  bought  at  each  price. 

16.  From  two  kinds  of  coffee  selling  at  30  cents  and  35  cents, 
respectively,  a  grocer  wishes  to  get  a  mixture  of  20  pounds  to 
be  sold  at  32  cents  a  pound.  How  many  pounds  of  each  kind 
must  he  use  ? 

17.  A  leaves  town  three  hours  before  B,  traveling  at  a  rate 
of  2^  mi.  an  hour.  B  travels  at  4  mi-  an  hour.  When  and 
where  does  B  overtake  A  ? 


28 


SECOND-YEAR  MATHEMATICS 


JlOl.  Elimination  by  comparison.  This  method  works 
well  if  one  of  the  unknowns  has  the  same  coefficient  in 
both  equations  of  the  system,  as  with 


4x=3 

4:x  =  5+2y 


The  value  of  x  to  be  determined  here  being  the  same 
for  both  equations,  it  follows  that  5+2t/=3.        Why? 
Hence,  y=—l. 


EXERCISES 


Solve  the  following  systems,  eliminating  by  comparison, 
doing  as  much  of  the  work  as  you  can  without  the  pencil. 


1. 


p+w=12 
p—w=—4i 

4. 

l+y=^ 

X     1    31/ 

l5~5"^  5 

x+y=3 
x  =  5-Sy 

x  =  niy-\-n^ 
x=  ny-\-rn?- 

6.  < 

f    x+  y  =  bm 
i    a;-37/  =  0 

PROBLEMS    LEADING    TO    EQUATIONS    IN    TWO    UNKNOWNS 

102.  Problems    about  work.     Solve    the    following, 
doing  all  you  can  orally : 

1.  If  the  time  required  to  do  a  piece  of  work  is  10  days,  what 
part  of  it  is  done  in  1  day  ?    In  3  days  ?    In  10  days  ? 

2.  If  the  entire  time  is  x  days,  what  part  of  the  work  is 
done  in  1  day  ?    In  3  days  ?    In  a:  days  ? 


METHODS  OF  ELIMINATION  29 

3.  If  the  time  is  y  days,  what  part  of  the  work  is  done  in 
1  day  ?    In  3  days  ?    Iny  days  ? 

4.  If  A  works  3  days  on  a  piece  of  work  and  B,  2  days,  they 
do  TT  of  it.  But  if  A  works  2  days  and  B,  3  days,  they  do  yV 
of  it.    In  how  many  days  could  each  one  do  it,  working  alone  ? 

Letting  x  and  y  denote  the  number  of  days  required  by  A  and 
B,  respectively,  then  -  and  -  will  denote  the  parts  A  and  B,  respec- 
tively, can  do  in  one  day. 

Whence,        -+-  =  ^  (1) 

X    y     lb 

and,  -+-  =  :^  (2) 

'  X     y     10  ^  ^ 

These  equations  are  not  linear  in  x  and  y,  but  are  linear  in  - 

and  -.     They  should  not  be  cleared  of  fractions,  but  -  or  -  should 
y  '         X       y 

first  be  eliminated,  thus. 


x^y 

28 
"l5 

x^y 

27 
^10 

Subtracting, 

_5 

y~ 

25 
30 

Whence, 

1 

y~ 

1 
^6 

and,  ?/  =  6 

Substituting  in  equation  (1),        x  =  b. 


Solution: 


jx  =  5 
\y  =  Q 


30 


SECOND-YEAR  MATHEMATICS 


EXERCISES 


Solve  the  following  systems  of  equations  without  clearing 
of  fractions,  and  check  them: 


2. 


3. 


4. 


x^y     15 

1     1      2 

.  X    y~15 

t5.  < 

12    25    „ 

9.  ^ 

x^y       12 
I  X    y 

X    y 
I  X    y 

6. 

X    y 

IP. 

1  ^^  - 

10    9      1 

X     y~20 

I  x^y    6 

17. 

[   5    6       24 
a;  "^2/  "143 
13     11      1 
.  X      y  "30 

111.  < 

11     2       1 

X     y  "6 
2    3      31 

I  x^y  "24 

7    4    11 
x^y~SO 
5    6         3 

[  X    y~     28 

t8.  - 

7    9  _  22 

x'^y      105 

15    21         4 

.X      y~     21 

U2.  < 

6    3      13 
x^y  760 
9    20     7 

.   x^y~12 

MISCELLANEOUS   PROBLEMS 

103.  Solve  the  following  problems: 

1.  A  boy  is  17  months  5  days  older  than  his  sister.  After 
21  days  he  is  twice  as  old  as  his  sister.     How  old  is  each  ? 

2.  Two  kinds  of  coffee,  one  at  32  cents  a  pound,  the  other  at 
25  cents  a  pound  are  to  be  mixed  in  the  ratio  3 : 2.  How  many 
pounds  of  each  must  be  taken  to  make  a  mixture  to  cost  $8 .  00  ? 

3.  The  sides  of  a  rectangle  are  to  each  other  as  3:8.  Find 
the  lengths  of  the  sides  if  the  perimeter  is  283. 

4.  Two  sums  are  invested  at  3  per  cent  and  3^  per  cent, 
respectively,  bringing  an  annual  income  of  $52 .  60.  If  the  first 
sum  is  invested  at  3|  per  cent  and  the  second  at  3  per  cent,  the 
annual  income  is  $52 .  70.    What  are  the  two  sums  ? 


METHODS  OF  ELIMINATION  31 

5.  Three  times  the  reciprocal  of  the  first  of  two  numbers 
and  4  times  the  reciprocal  of  the  second  are  together  equal  to  5. 
Seven  times  the  reciprocal  of  the  first  less  6  times  the  reciprocal 
of  the  second  is  equal  to  4.     What  are  the  numbers  ? 

Summary 

104.  In  this  chapter  the  processes  of  solving  linear 
equations  in  two  unknowns  graphically  and  by  eliminating 
magnitudes  have  been  extended. 

The  following  processes  have  been  studied  for  the  first 
time: 

(1)  Elimination  by  substitution. 

(2)  Elimination  by  comparison. 


CHAPTER  IV 


QUADRILATERALS.    PRISMATIC  SURFACE. 
DIEDRAL  ANGLES 

Parallelograms 

105.  Parallelogram.     A  quadrilateral  having  both  pairs 

of  opposite  sides  parallel  is  a  parallelo-        ^ 

gram,     (See  Fig.  23.)  / / 

106.  Uses  of  the  parallelogram.        Fig.  23 
Some   designs  are  based  upon  the 

parallelogram.  A  designer  some- 
times constructs  tile  flooring,  Fig.  24, 
from  a  network  of  parallelograms. 
Give  other  examples  of  designs 
based  upon  the  parallelogram. 

Tops  of  desks  and  tables, 
blackboards,  windows,  walls,  picture 
frames,  etc.,  are  examples  of 
parallelograms. 

Constructions  with  the  parallel 
ruler,  Fig.  25,  which  is  used  to  draw  parallel  lines,  are 
based  upon  a  property  of  parallelograms.     (See  §  124). 


Fig.  24 


Fig.  25 


Fig.  26 


The  same  principle  is  used  in  the  construction  of  the 
adjustable  shelf.  Fig.  26,  which  remains  in  horizontal 
position  as  it  is  moved  to  and  from  the  wall. 

32 


QUADRILATERALS 


33 


Fig.  27 


Fig.  28 


Surveyors  make  use  of  a  property  of  the  parallelogram 
to  lay  off  parallel  lines,  Fig.  27,  or  to  extend  a  line  beyond 
an   obstacle. 
Fig.  28  (see 
§  125). 

The  use  of 
the  parallelo- 
gram   in    physics    may    be    seen    from    the    following 
problem : 

The  wind  drives  a  steamer  northeastward  with  a  force  which 
would  carry  it  12  miles  per  hour,  and  the  engine  is  driving  it 
southward  with  a  force  which  would  carry  it  15  miles  per  hour. 
What  distance  will  it  travel  in  an  hour  and  in  what  direction  ? 

Let  AB,  Fig.  29,  represent  in  magnitude 
and  direction  the  12-mile  rate  toward  the 
northeast  and  AC  the  15-mile  rate  south- 
ward; then  it  is  shown  by  experiments  that 
the  rate  and  direction  in  which  the  boat 
actually  moves  may  be  represented  by  a  line- 
segment  as  follows : 

Construct  a  parallelogram  as  ABDC  on 
AB  and  ^C  as  adjacent  sides  and  draw  a 
Hne-segment  from  A  to  the  opposite  vertex, 
D.  The  diagonal  Une  AD  is  the  required 
segment.  Before  we  can  solve  this  problem 
we  must  know  how  to  construct  the  parallelogram  from  these 
given  parts. 

107.  Construction  of  parallelograms.  To  construct  a 
parallelogram  having  given  two  adjacent  sides  and  the 
included  angle. 

Given  two  segments,  a  and  b  and  angle  x,  Fig.  30. 

Required  to  construct  a  parallelogram  having  two 
adjacent  sides  equal  to  a  and  6,  and  including  an  angle 
equal  to  x. 


Fig.  29 


34  SECOND-YEAR  MATHEMATICS 

Construction:  Suppose  a  =  1 . 5  in.,  6  =  1  in.,  and  x  =  45' 


b  =i 


a  »  1.5 


C 

Fig.  30 


Draw  a  line  as  AB. 

On  A5,  lay  off  AC  =  a.  '  ^ 

At  A,  construct  line  AD  making  with  AB  angle  x 

equal  to  angle  x. 

On  AD  lay  off  A^  =  6. 

With  C  as  center  and  radius  equal  to  h,  draw  arc  1. 

With  E  as  center  and  radius  equal  to  a,  draw  arc  2 
intersecting  arc  1  at  F. 

Draw  £;F  and  CF. 

ACFE  is  the  required  parallelogram. 

The  proof  that  ACFE  is  a  parallelogram  is  based  upon 
one  of  the  properties  of  parallelograms  to  be  studied 
later  in  this  chapter  (see  §  124). 

EXERCISES 

1.  Construct  a  parallelogram  having  a  =  2  in.,  6  =  1.5  in., 
and  x  =  50°,  and  compare  it  with  the  parallelograms  constructed 
by  other  members  of  the  class. 

2.  How  do  two  parallelograms,  having  two  sides  and  the 
included  angle  equal  respectively,  seem  to  compare  in  size  and 
shape  ? 

3  Prove  that  two  parallelograms  are  congruent,  if  two 
adjacent  sides  and  the  included  angle  of  one  are  equal,  respectively, 
to  the  corresponding  parts  of  the  other. 

Proof  hy  superposition:  Apply  one  of  the  paraUelograms  to  the 
other  and  show  that  they  can  be  made  to  comcide  throughout. 


QUADRILATERALS  35 

4»  On  squared  paper  construct  the  parallelogram  of  the 
steamer  problem  in  §  106,  and  find  the  solution  by  measuring 
the  diagonal  AD,  and  ZCAD,  Fig.  29. 

5.  Represent  graphically  a  force  of  20  lb.  acting  northeast 
and  a  force  of  30  lb.  acting  northwest  upon  the  same  body. 
What  single  force  has  the  same  effect  upon  the  motion  of  the 
body  as  the  two  forces  together?  In  what  direction  will  the 
body  move  ? 

108.  Before  taking  up  the  study  of  the  parallelogram, 
we  shall  recall  some  facts  about  parallel  lines.  At  the 
same  time  we  shall  discuss  and  exemplify  two  important 
methods  of  proof,  that  were  not  given  in  chapter  II. 

109.  Indirect  method  of  proof.  The  indirect  method 
may  be  illustrated  by  proving  the  following  theorem: 

Theorem:  //  each  of  two  lines  is  parallel  to  a  third 
line,  they  are  parallel  to  each  other. 

Given  AB\\  EF,    CD\\  EF,       c 1 

Fig.  31.  -E F 

To  prove  AB\\  CD.  F^«- 31 

Proof:  Assume  AB  not  parallel  to  CD. 

Then  AB  and  CD  intersect  at  some  point,  as  P,  if 
far  enough  extended.  For,  two  intersecting  lines  have  one 
point  in  common  (§4). 

But  PA  II  FE,  by  hypothesis, 

and  PC  II  FE,  by  hypothesis. 

Hence,  there  are  two  lines  parallel  to  FE  passing 
through  the  point  P. 

This  is  impossible.  For,  through  a  point  outside  of 
a  given  line  only  one  line  can  be  drawn  parallel  to  the 
given  line  (§41). 

Therefore  the  assumption  that  AB  is  not  parallel 
to  CD  is  wrong,  and  AB  II  CD. 


36  SECOND-YEAR  MATHEMATICS 

It  is  thus  seen  that  the  indirect  method  of  proof  con- 
sists of  the  foUowing/oi^r  steps,  numbered  I,  II,  III,  and  IV: 

I.  Make  an  assumption  which  denies  the  conclusion  of 
the  theorem. 

Thus,  if  you  are  to  prove  that  a  =  h,  assume  that 
a9^h,  or  if  A B  is  to  be  proved  parallel  to  CD,  assume  AB 
not  parallel  to  CD. 

II.  By  correct  reasoning  show  that  the  assumption  leads 
to  an  absurdity. 

In  the  preceding  theorem  the  absurdity  is  the  state- 
ment that  two  lines  can  be  drawn  parallel  to  the  same  line 
passing  through  a  given  point. 

III.  It  then  follows  that  the  assumption  is  wrong. 
For,  if  we  start  right,  correct  reasoning  cannot  lead 

us  to  a  wrong  conclusion.  To  reach  a  correct  conclusion 
in  a  course  of  reasoning,  two  things  are  necessary,  and  only 
two,  namely:  (1)  the  premises  from  which  the  reasoning 
starts — in  geometry,  we  call  them  the  assumptions — must 
he  correct,  and  (2)  the  reasoning  must  he  sound. 

If  a  certain  conclusion  is  known  to  be  incorrect,  the 
assumption  from  which  the  reasoning  starts  is  incorrect 
or  the  reasoning  is  faulty,  or  both.  If  a  conclusion  is 
incorrect  and  the  reasoning  is  sound,  the  assumption  must 
he  incorrect. 

IV.  Hence,  the  conclusion  is  correct  and  the  theorem  is 
proved. 

110.  Theorem:  Two  lines  that  are  perpendicular 
to  the  same  line  are  parallel. 

Given    AB±EF,    CDLEF,  ^ 

Fig.  32.  A- 

To  prove  A  5  II  CD. 


Proof  (indirect  method) :  f 

Assume  AB  not  parallel  to  CD.  Fig.  32 


QUADRILATERALS  37 

Then  AB  and  CD  intersect  at  some  point,  P.       Why  ? 

Hence,  PA  ±  EF  and  PC  ±  EF.        Why  ? 

This  is  impossible.         Why  ? 

Therefore,  the  assumption  that 
AP  is  not  parallel  to  CD  is  wrong, 
and  AB  \\  CD. 

EXERCISES 

1.  Show  that  this  theorem  affords  Fiq.  33 
a  very  simple   way  of  drawing 

parallel  lines  by  means  of  a  T-square  (Fig.  33). 

2.  State  the  conditions  which  make  two  lines  parallel 
to  each  other. 

3.  Draw  two  parallel  Hues,  using  §  110. 

4.  Point  out  in  the  classroom  two  lines  not  in  the 
saitie  plane,  but  perpendicular  to  the  same  Hne.  Are  these 
Knes  parallel  ? 

111.  The  method  of  proof  used  in  §§  109-110  is  com- 
monly known  as  a  reductio  ad  ahsurdum*  or  a  reduction 
to  an  absurdity,  which  means  that  from  assuming  the  nega- 
tion of  the  conclusion  of  the  theorems  in  question  we  are 
(by  correct  reasoning)  led  to  a  statement  which  is  contra- 
dictory to  known,  or  accepted,  facts.  It  is  a  powerful 
method  of  proof,  used  not  only  in  geometry  but  in  everyday 
life. 

*Eudoxus  of  Cnidus  (408  B.C.),  founder  of  the  School  at 
Cyzicus,  and  a  contemporary  of  Plato,  used  the  reductio  ad  ahsurdum 
method.  In  his  Short  Account  of  the  History  of  Mathematics  (5th  ed.) , 
Ball  says  (p.  39)  that  while  the  principle  of  the  reductio  ad  ahsurdum 
had  been  used  occasionally  before,  Hippocrates  of  Chios  (b.  about 
470  B.C.)  drew  attention  to  it  as  a  legitimate  mode  of  proof,  capable 
of  numerous  applications.  In  this  sense  Hippocrates  may  be  re- 
garded as  having  introduced  the  method. 


38  SECOND-YEAR  MATHEMATICS 

112.  Method  of  analysis.*  The  following  example 
will  illustrate  the  method  of  analysis: 

Theorem:  //  two  alternate  interior  angles,  formed  by 
two  lines  and  a  transversal  are  equal,  the  lines  are 
parallel. 

Given  A  B,  CD,  and  the  transversal  EF;  a=a\  Fig.  34. 

To  prove  AB\\  CD. 

Preliminary  discussion: 
To    prove    AB\\CD,    we 
may  begin  by  asking  the     ^ — — 
general  question:    When  Fig.  34 

are  two  lines  parallel  ? 

Thus  we  know  that  AB  is  parallel  to  CD,  if  both  lines 
are  perpendicular  to  the  same  line  (§  110). 

This  suggests  drawing  a  line,  as  GH,  perpendicular 
to  one  of  the  given  lines  and  then  proving  it  to  be  per- 
pendicular to  the  other. 

GH  is  perpendicular  to  AB,ii  Z  GHF  is  a  right  angle. 

We  may  show  Z  GHF  to  be  a  right  angle  by  showing 
that  it  is  equal  to  the  right  angle  HGE. 

This  will  be  true,  if  we  can  show  AMHF^  AMGE. 

In  triangles  MHF  and  MGE,  we  know  that  x  =  x' 
and  a  =  a\  which  is  not  sufficient  to  make  the  triangles 
congruent.  But  by  taking  M,  so  that  EM  =  MF,  we 
will  have  the  third  part  which  is  necessary  to  make 
AMHF^AMGE. 

Being  able  to  prove  AMHF^AMGE,  it  may  be 
possible  so  to  reverse  the  steps  in  this  discussion  as  to 
prove  the  lines  AB  and  CD  to  be  parallel.  This  may  be 
done  as  follows: 

*  Plato  (429-348  B.C.)  is  said  to  have  formulated  this  method 
of  proof. 


QUADRILATERALS  39 

Proof:  Bisect  EF  Sit  M. 

Through  the  middle  point,   M,   of  EF,   draw  MG 
perpendicular  to  CD  and  prolong  GM  to  meet  AB  Sit  H. 
Prove  that  A  EGM  ^  A  MHF  (a.s.a.,  §  67) . 

y  =  y^  Why? 

But,  y  =  Tt.  /.     Why? 

.-.     2/'  =  rt.  Z     Why? 
.'.  AB  and  CD  are  both  perpendicular 

toG^     Why? 
.-.ABWCD       Why? 

It  is  seen  that  the  method  of  analysis  consists  of  the 
following  four  steps : 

I.  Ask  the  question:  "Under  what  conditions  is  the 
conclusion  true?"  Select  from  the  answers  the  one  you 
think  you  can  establish  to  be  true.  Thus,  when  AB  is 
to  be  proved  parallel  to  CD  the  question  should  be, 
''When  are  two  lines  parallel?"  The  answer  will  be: 
''We  have  proved  previously  that  two  lines  are  parallel, 
(1)  if  they  are  perpendicular  to  the  same  line,  or  (2)  if 
they  are  parallel  to  the  same  line."  From  these  two 
possibilities,  select  the  one  you  think  you  can  prove  to 
be  true.  The  conclusion  is  true,  if  the  truth  of  this 
second  fact  is  established. 

II.  Repeat  the  same  process  of  reasoning  with  the 
second  fact,  thus:  This  second  fact  is  true,  if  a  certain 
third  fact  can  be  proved. 

III.  Continue  this  type  of  reasoning  until  you  deduce 
a  fact  that  is  known  to  be  true. 

IV.  Starting  from  this  known  fact,  reverse  the  pro- 
cess, proving  every  statement,  until  the  conclusion  is 
reached. 

113.  Proof  by  analysis.     Step  IV,  §  112,  is  the  proof 

of   the   theorem.     The   preliminary   reasoning   in   steps 


dO  SECOND-YEAR  MATHEMATICS 

I,  II,  and  III  is  called  the  analysis.  The  purpose  of 
the  analysis  is  to  enable  the  student  to  discover  the 
known  fact  from  which  to  start,  and  to  learn  how  to 
arrange  the  proof.  In  the  demonstration  of  a  theorem 
only  the  proof  is  given. 

114.  Converse  of  a  theorem.  A  theorem  is  said  to 
be  the  converse  of  another  theorem,  if  the  hypothesis 
and  conclusion  of  one  are,  respectively,  the  conclusion 
and  hypothesis  of  the  other. 

State  the  converse  of  the  following: 

1.  If  two  sides  of  a  triangle  are  equal,  the  angles  opposite 
them  are  equal. 

2.  In  a  circle  equal  arcs  are  subtended  hy  equal  chords. 

Are  the  converses  of  the  following  statements  true  ? 

If  two  angles  are  right  angles  they  are  equal. 

If  two  parallelograms  have  equal  bases  and  altitudes, 
they  are  equal. 

All  righteous  people  are  happy. 

If  two  angles  of  a  triangle  are  equal  the  sides  opposite 
them  are  equal. 

Thus,  because  a  theorem  is  true,  it  does  not  follow 
that  the  converse  is  true.  Since  some  converses  are  true 
and  some  are  not,  a  proof  is  necessary  before  the  converse 
can  be  accepted  as  true. 

115.  Methods  used  to  prove  the  converse  of  a  theorem. 

Two  methods  are  used  most  frequently  to  test  the  truth 
of  the  converse  of  a  theorem. 

1.  If  the  steps  of  the  proof  of  the  original  theorem 
are  reversible,  use  this  proof  as  analysis  and  retrace  it, 
step  by  step,  until  the  hypothesis  is  reached.  Since  the 
hypothesis  of  the  original  theorem  is  the  conclusion  of 
the  converse,  this  proves  the  converse. 


QUADRILATERALS  41 

2.  The  indirect  method. 

116.  Theorem:  //  two  parallel  lines  are  cut  by  a  trans- 
versal,  the  alternate  interior  angles  are  equal.  (C9nverse 
of  the  theorem  in  §  112.) 

Given    AB  II  CD.     AB 
and    CD    cut    by    EH,     _^  ^.^Z----" 'g 

Fig.  35.  G-'-"""" 

To  prove  a=a'. 

c / 

Proof  (indirect  method) :  /^ 

Suppose  a  9^  a'  Fig.  35 

Draw  GF  making  h  =  a'. 

Then,  GF  II  CD        Why? 

But,  ABWCD        Why? 

It  is  impossible  that  both  GF  and  AB  are  parallel  to 
CD.     Why? 

Therefore,  the  assumption  that  a  9^  a'  is  wrong,  and 
a  =  a\ 

EXERCISE 

Prove  that  if  one  of  two  parallel  lines  is  perpendicular  to  a 
third  line  the  other  is  also. 

117.  Properties  of  parallelograms.     In  §  106  it  was 

seen  that  some  of  the  properties  of  parallelograms  could 
be  applied  in  a  number  of  ways.  We  will  now  prove  the 
following : 

If  a  quadrilateral  is  a  parallelogram  — 

1.  A  diagonal  divides  it  into  congruent  triangles; 

2.  The  opposite  sides  are  equal; 

3.  The  opposite  angles  are  equal; 

4.  The  consecutive  angles  are  supplementary; 

5.  The  diagonals  bisect  each  other. 


42 


SECOND-YEAR  MATHEMATICS 


118.  Theorem:  A  diagonal  divides  a  parallelogram 
into  two  congruent  triangles. 

Given  the  parallelogram  A  B  CD 
with  thie  diagonal  A  C,  Fig.  36. 

To  prove  that 

AABC^AADC. 

Analysis:  What  conditions  are  sufficient  to  make 
two  triangles  congruent?  In  what  ways  can  we  prove 
two  angles  equal,  and  which  of  them  may  be  used  to 
prove  x  =  x'f 


Fig.  36 


Proof: 

STATEMENTS 

REASONS 

AC=AC 

Common 

DC  II  AB 

Since  ABCD  is  a  parallelo- 

gram by  hypothesis. 

,\x  =  x' 

If  two  parallel  Hues  are  cut 

by  a  transversal  the  alter- 

nate interior  angles  are 

equal. 

AD  II  BC 

By  hypothesis. 

-'-y^y' 

Alternate    interior    angles 

formed  by  parallel  lines 

and    a    transversal    are 

equal. 

AABC^AADC 

a.s.a. 

119.  Theorem:    The  opposite  sides  of  a  parallelogram 
are  equal  (method  of  congruent  triangles). 

Use  §  118. 

120.  Theorem:   The  opposite  angles  of  a  parallelogram 
are  equal. 

Use  §  118  to  prove  ZD=ZB,  Fig.  36.     Then  draw 
diagonal  DB  to  prove  Z.A=  ZC. 


QUADRILATERALS     .  43 

121.  Theorem:  The  consecutive  angles  of  a  parallelo- 
gram are  supplementary. 

Notice  that  the  consecutive  angles  are  interior  angles 
on  the  same  side,  formed  by  two  parallels  cut  by  a  trans- 
versal.    Use  §  47. 

122.  Theorem:  The  diagonals  of  a  parallelogram 
bisect  each  other.  Dyr^ ^c 

Given  the  parallelogram  A  B  CD  /    "^>^--''''   / 

with  the  diagonals  AC  and  BD,        ZillI^l^Jjlli^ 


Fig.  37.  Fig.  37 

To  prove  that  AE  =  EC,  DE  =  EB. 

Analysis:  How  may  two  line-segments  be  proved 
equal  ? 

Which  of  these  ways  seems  the  most  promising  to 
^TOYeAE  =  ECf 

Proof:  Prove  AD^C^AA^^. 

AE  then  equals  EC,  and  BE  equals  ED.    Why  ? 

EXERCISES 

1.  Prove  that  if  one  of  the  angles  of  a  parallelogram  is  a 
right  angle,  all  the  angles  are  right  angles. 

2.  Prove  that  if  two  adjacent  sides  of  a  parallelogram  are 
equal,  all  the  sides  are  equal. 

3.  Prove  that  parallels  intercepted  between  parallels  are 
equal. 

4.  Prove  that  parallels  are  everywhere  equally  distant. 

5.  One  pair  of  opposite  sides  of  a  parallelogram  is  denoted 
by  x^-\-z  and  6(3— x)  and  the  other  pair  by  y^—y  and  3(5-?/). 
Find  X  and  y,  and  the  lengths  of  the  sides. 

t6.  Two  opposite  angles  of  a  parallelogram  are  denoted  by 
a;' +6  and  7{x-\-2).  Find  x  and  all  the  angles  of  the  parallelo- 
gram. 


44  SECOND-YEAR  MATHEMATICS 

123.  Conditions  under  which  a  quadrilateral  is  a 
parallelogram. 

In  the  following  it  will  he  proved  that  a  quadrilateral  is  a 
parallelogram — 

1.  If  the  opposite  sides  are  parallel; 

2.  If  the  opposite  sides  are  equal; 

3.  If  one  pair  of  opposite  sides  are  equal  and  parallel; 

4.  If  the  opposite  angles  are  equal; 

5.  //  the  diagonals  bisect  each  other. 

124.  Theorem :  If  the  opposite  sides  of  a  quadrilateral 
are  equal,  the  quadrilateral  is  a  parallelogram. 

Given  the  quadrilateral  ABCD,  Fig.  38,  having 
AB  =  DC,  AD  =  BC. 

To      prove      AB  ||  DC, 
AD  II  BC. 

Proof:  Draw  AC. 

Prove    AABC^AADC.  ^     _ 

(s.s.s.)  ^'^-^^ 

Then  x  =  x\  and  y  =  y'         Why  ? 

Hence,  AB  ||  DC  and  AD  II  BC.        Why  ? 

125.  Theorem:  //  one  pair  of  opposite  sides  of  a 
quadrilateral  are  equal  and  parallel,  the  quadrilateral  is  a 
parallelogram. 

Given  the  quadrilateral  ABCD,  Fig.  39,  having 
AB  =  DC,  and  ABWDC. 

To  prove  that  ABCD  is  a  paral- 
lelogram. * 

Proof:  Prove  AABC^AADC. 
(s.a.s).  Fig.  39 

Then  AD  =BC.        Why? 

Use  the  theorem  of  §  124  to  prove  that  ABCD  is  a 
parallelogram. 


QUADRILATERALS  45 

126.  The  proof  of  the  following  theorem  is  a  good 
example  of  the  algebraic  method  of  proof: 

Theorem:   If  the  opposite  angles  of  a  quadrilateral  are 
equal,  the  quadrilateral  is  a  parallelogram. 

Given  the  quadrilateral  A  BCD, 
Fig.  40,  having  a  =  c,b  =  d. 

To  prove  AB  \\  DC,  AD  II  BC. 

Analysis:     Under    what    con-     ""  yiq.  40 

ditions    are    two    lines    parallel? 

What  relations  are  known  between,  a,  h,  c,  and  d  f  How 
may  we  obtain  from  these  relations  a  relation  which  will 
show  that  AJ5  II  DC  f 

Proof: 

STATEMENTS  REASONS 

a+6-fc-fd  =  360  Why? 

a  =  c  Why? 

h  =  d  Why? 

Hence,  a+cZ+a-f  d  =  360  By  eliminating  b  and  c. 

2a+2d  =  360  Combining  like  terms. 

a-\-d=  ISO  Why? 

Hence,  AB  \\  DC  Why? 

Similarly,  prove  that  AD  ||  BC. 

127.  //  the  diagonals  of  a  quadrilateral  bisect  each  other, 

the  quadrilateral  is  a  parallelogram. 

D  c 

Draw  the  diagonals  AC  and  BD, 

Fig.  41. 

Prove  ADEC^AAEB. 

Then  DC  =  AB. 

Similarly,  show  that  AD  =  BC. 

Hence,  ABCD  is  a  parallelogram.    Why  ? 


v^ 

^/7/\ 

\     ^^ 

■^   r. 

yy\ 

\ 

\ 

\^. 

-  X' 

A 

B 

FlQ. 

41 

46 


SECOND-YEAR  MATHEMATICS 


128.  Classification  of  quadrilaterals.     Quadrilaterals 
may  be  classified  as  follows : 

Parallelogram.     A   quadrilateral    having    two    pairs    of 

opposite  sides  parallel  is  a  parallelogram. 
Rhomboid.     A   parallelogram   whose  angles  are  oblique 

is  a  rhomboid. 
Rhombus.     An  equilateral  rhomboid  is  a  rhombus. 
Rectangle.     A    parallelogram    whose    angles    are    right 

angles  is  a  rectangle. 
Square.    An  equilateral  rectangle  is  a  square. 
Trapezoid.     A  quadrilateral  having  one  pair  of  opposite 

sides  parallel  is  a  trapezoid. 
Isosceles  trapezoid.     If  the  two  non-parallel  sides  are 

equal  the  trapezoid  is  isosceles.     The  parallel  sides 

of  the  trapezoid  are  the  bases. 

The  same  classification  is  represented  in  the  following 
table: 


Quadrilateral 
(figure  of 
4  sides) 


Parallelogram 

(2  pairs  of 

opposite  sides 

parallel) 


Trapezoid 

(1  ijairof 

opposite  sides 

parallel) 

/       \ 


Rhomboid 

(oblique  angles, 

consecutive  sides 

unequal) 


\ZZ\ 


Rectangle 

(right  angles, 

consecutive  sides 

unequal) 


Rhombus 

(oblique  angles. 

equilateral) 


Square 
(right  angles, 
equilateral) 


Isosceles  trapezoid 

(2  non-parallel 

sides  equal ) 


QUADRILATERALS 


d7 


EXERCISES 


Prove  the  following: 

1.  If  two  consecutive  sides  of  a  rectangle  are  equal,  all  the 
sides  are  equal. 

2.  If  the  diagonals  of  a  parallelogram  are  equal,  the  figure 
is  a  rectangle. 

3.  The  diagonals  of  a  rectangle  are  equal. 

4.  The  diagonals  of  a  square  are  equal. 

5.  The  diagonals  of  a  rhombus  bisect  each  other  perpen- 
dicularly. 

6.  The  diagonals  of  a  square  bisect  each  other  perpen- 
dicularly. 

7.  If  the  diagonals  of  a  parallelogram  bisect  each  other 
perpendicularly,  the  figure  is  a  rhombus,  or  a  square. 

8.  A  circle  may  be  circumscribed  about  a  rectangle,  or  a 
square. 

9.  If  the  angles  of  a  parallelogram  are  bisected  by  the 
diagonals,  the  figure  is  a  rhombus,  or 
a  square. 

tlO.  If  the  midpoints  of  two  oppo- 
site sides  of  a  parallelogram  are  joined 
to  a  pair  of  opposite  vertices.  Fig.  42, 
a  parallelogram  is  formed. 

|11.  In  the  parallelogram.  Fig.  43, 
AE=BF  =  CG  =  DH.  Prove  that 
EFGH  is  a  parallelogram. 

$12.  The  perpendiculars  to  a 
diagonal  of  a  parallelogram  from  the 
vertices  not  on  the  diagonal  are  equal. 
Fig.  44,  i.e.,  DE  =  BF. 

13.  If  two  points  on  the  same  side 
of  a  fine  are  equally  distant  from  the 
line,  the  fine  passing  through  the  two 
points  is  parallel  to  the  given  line. 


; 

E 

^4 

,^ 

1^^ 

Fig.  .42 

D^ 

G 

"f 

-^  V 

A 

--^: 

'^        E 

B 

FiQ.  43 

D^ 

/ 

v,--^ 

A 

V  V 

Fig.  44 


48 


SECOND-YEAR  MATHEMATICS 


1 14.  The  bisectors  of  two  opposite  angles  of  a  parallelogram 
are  parallel. 

1 15.  The  bisectors  of  the  angles  of  a 
parallelogram  form  a  rectangle. 

J 16.  The  bisectors  of  the  angles  of  a 
rectangle  form  a  square. 

Jl7.  The  sum  of  the  perpendiculars  from 
a  point  on  the  base  of  an  isosceles  triangle  to 
the  two  equal  sides  is  equal  to  the  altitude  to 
one  of  these  sides.    Fig.  45. 

1 18.  The  sum  of  the  perpendiculars  from 
a  point  within  an  equilateral  triangle  to  the 
three  sides  is  equal  to  the  altitude.    Fig.  46.  Fig.  46 


FiQ.  45 


Constructions 
129.  Make  the  following  constructions : 

1.  Given  a  side  and  the  diagonal  of  a  rectangle,  construct 
the  rectangle. 

2.  Given  a  side  and  an  angle  of  a  rhombus,  construct  the 
rhombus. 

3.  Given  the  diagonal  of  a  square,  construct  the  square. 

4.  Given  the  diagonals  of  a  rhombus,  construct  the  rhombus. 


Problems 

130.  Solve  the  following  problems  algebraically: 

1.  The  diagonals  of  a  rectangle  are  denoted  by  x^—x  and 
2(2x+7).    Find  both  values  of  x  and  the  diagonals. 

2.  The  diagonals  of  a  parallelogram  divide  each  other  so 
that  the  segments  of  one  are  x^-{-x  and  2(5x— 7),  and  of  the 
other,  t'^+2t  and  8(3—0.  Find  x,  t,  and  the  lengths  of  the 
diagonals. 


QUADRILATERALS 


49 


|3.  The  diagonals  of  a  rhomLas  divide  each  other  so  that 
the  parts  of  one  diagonal  are  denoted  by  a;^  and  3  (2a: +9),  and 
of  the  other  by  if  and  2(^+4).  Find  x,  y,  and  both  of  the 
diagonals. 

|4.  Two  of  the  four  angles  that  the  diagonals  of  a  rhombus 
make  with  each  other  are  given  by  a;^— 10  and  10(2a;  — 11). 
Find  X  and  the  four  angles. 


Quadratic  Equations 

131.  Solve  the  following  equations,  using  either  the 
method  by  factoring  or  by  completing  the  square: 


1.  a;2  =  5x-4 

2.  a;2+l  =  2(a;+18) 


3.  x2-a;  =  3x-4 

4.  x'^-4:  =  x+lQ 


The  Trapezoid 

132.  Prove  the  following: 

1.  If  the  two  angles  at  the  ends  of  a  base  of  a  trapezoid  are 
equal,  the  trapezoid  is  isosceles. 


E 

Fig.  47 


Draw  CE  \\  DA,  Fig.  47. 
Prove  C^  =  C5. 

2.  If  the  non-parallel  sides  of 
a  trapezoid  are  equal,  the  angles 
at  the  ends  of  a  base  are  equal. 

J3.  Prove  that  the  diagonals  of  an  isosceles  trapezoid  are 
equal.  j^ 


The  Rite 

133.  The  kite.  A  quadrilateral 
having  two  pairs  of  adjacent  sides 
equal,  is  a  kite. 

Thus,  ABCD,  Fig.  48,  is  a  kite 
if  AD==DC,  and  AB  =  BC. 


50 


SECOND-YEAR  MATHEMATICS 


EXERCISES 

Prove  the  following: 

1.  The  diagonals  of  a  kite  are  perpendicular  to  each  other. 

2.  One  pair  of  opposite  angles  of  .a  kite  are  equal,  i.e., 

ZA=ZC. 

« 

Sjnnmetry 

134.  Axis  of  symmetry.  Aline 
is  called  an  axis  of  syimnetry  of  a 
figure  if  it  is  the  perpendicular 
bisector  of  all  line-segments  joining 
corresponding  points  of  the  figure. 
Thus,  AE,  Fig.  49,  is  the  axis  of 
symmetry  in  DCBAB'C'D'. 


EXERCISES 

1.  What  is  the  axis  of  symmetry  of  a  fine-segment  ? 

2.  Draw  the  axis  of  symmetry  of  a  given  angle. 

3.  Draw  an  axis  of  symmetry  in  a  given  equilateral  triangle. 
Prove  the  following: 

4.  The  diagonal  BD  of  the  kite, 
Fig.  50,  is  an  axis  of  symmetry. 

5.  The  point  of  intersection  E  of  the 
diagonal  D5,  Fig.  50,  and  the  bisector  of 
ZA  (axis  of  symmetry  of  ZA),  is  equi- 
distant from  AD  and  AB  and  therefore 
may  be  taken  as  a  center  of  a  circle 
inscribed  in  the  kite. 

Use  Exercise  4. 

6.  The  perpendicular  bisector  of  a  base  of  an  isosceles  trape- 
zoid is  an  axis  of  symmetry. 

Prove  by  superposition. 


QUADRILATERALS 


51 


7.  The  point  of  intersection  of  the  perpendicular  bisectors 
of  one  of  the  bases  and  of  one  of  the  non-parallel  sides  of  an 
isosceles  trapezoid  is  equidistant  from  the  4  vertices.  Hence, 
a  circle  can  be  circumscribed  about  an  isosceles  trapezoid. 

8.  Each  diagonal  of  a  rhombus  is  an  axis  of  symmetry, 
Hence,  a  circle  can  be  inscribed  in  any  rhombus. 

Loci* 

135.  Solve  the  following  problems: 

1.  Where  must  the  center  of  a  wheel  lie  while  the  wheel 
rolls  along  a  straight  track  ? 

2.  Find  the  place  (locus)  of  a  point  in  a  plane  having  a 
fixed  distance  from  a  given  line. 

3.  Find  the  locus  of  a  point  in  a  plane  equidistant  from 
two  parallel  lines. 

Surfaces 

136.  Prismatic  surface.  Prism.  Given  a  polygon 
ABCD  .  .  .  ,  Fig.  51.     A  straight  line  AA',  not  in  the 


Fig.  51 


Fig.  52 


plane  of  the  polygon,  moves  always  remaining  parallel  to 
its  first  position  AA\  and  always  touching  the  polygon. 
A  A'  is  said  to  generate  a  prismatic  surface,  Fig.  52. 


*  Lod  is  the  plural  of  Iocils. 


52 


SECOND-YEAR  MATHEMATICS 


Let  the  polygon  ABCD  .... 
position,  A '5'C'2)' .  .  .  .  ,  Fig.  53, 
always  remaining  parallel  to  its  first 
position,  points  A,  B,  C,  .  .  .  . 
moving  along  the  straight  lines 
AA',  BE',  CC  .  .  .  ,  respectively. 
The  figure  thus  formed  is  a  prism. 

137.  Bases  of  prism.  Lateral 
surface.  The  parallel  polygons 
ABCD  ....  and  A'B'C'D'  .... 
are  the  bases  of  the  prism.  The 
portion  of  the  prismatic  surface 
between  the  bases  is  the  lateral  surface 


Fig.  52,  move,  to  a 


Fig.  53 


138.  Lateral  faces.  The  quadrilaterals  of  which  the 
lateral  surface  is  composed  are  the  lateral  faces  of  the 
prism. 

In  the  classroomi  point  out  a  prism  and  indicate  its  bases 
and  the  lateral  faces. 

EXERCISES 

1.  Show  that  the  lateral  faces  of  a  prism  are  parallelograms. 

2.  Show  how  to  generate 
according  to  the  method  of 
§  136  a  triangular  prism,  i.e.,  a 
prism  whose  base  is  a  triangle, 
Fig.  54. 

3.  Show  how  to  form  a 
parallelopiped  using  as  a  base 
a  parallelogram.  Fig.  55. 

4.  What  must  be  the  position  of  the  generating  line  AA', 
Fig.  55,  with  reference  to  the  plane  of  the  base  ABCD,  in  order 
that  all  the  lateral  faces  of  the  parallelopiped  be  rectangular  ? 


E'lQ.  54 


Fig.  55 


QUADRILATERALS  53 

Lines  and  Planes  in  Space 

139.  Determination  of  a  plane.  In  the  first-year 
course  the  pupil  has  become  acquainted  with  the  following 
important  solids  of  geometry:  the  cube,  parallelopiped, 
prism,  cone,  pyramid,  cylinder,  and  sphere  (§§203-13). 
In  the  study  of  these  solids,  he  has  learned  the  meaning 
of  such  terms  as  plane,  surface,  lines  perpendicular  to 
planes,  parallel  planes,  etc.  Illustrate  these  terms  on 
the  cube. 

The  pupil  has  seen  that  several  planes  may  pass 
through  (contain)  the  edge  of  a  solid,  or  through  two 
given  points.  Illustrate  this  with  a  cube,  or  by  using 
the  hinges  of  a  door  as  the  two  points  and  the  door  as  the 
plane. 

When  a  plane  passes  through  two  points  in  space, 
it  is  possible  to  let  the  plane  rotate  about  the  straight 
line  determined  by  these  points,  so  that  any  number  of 
planes  may  be  passed  through  the  line.  However,  the 
position  of  a  plane  is  fixed,  if  besides  making  it  pass 
through  a  given  straight  line,  we  make  it  pass  through  a 
fixed  point  not  on  the  given  line. 

The  conditions  which  determine  the  position  of  a  plane 
in  space  are  as  follows: 

1.  A  straight  line  and  a  point  not  in  that  line, 

2.  Three  points  not  in  the  same  straight  line. 

For,  if  two  of  the  points  are  joined  by  a  straight  line, 
condition  (1)  is  satisfied. 

3.  Two  intersecting  straight  lines. 

For  by  taking  one  of  the  lines  and  a  point  on  the 
other  (not  the  point  of  intersection),  condition  1  is  satis- 
fied. 


54  SECOND-YEAR  MATHEMATICS 

4.  Two  parallel  straight  lines. 

For  two  parallel  lines  lie  in  the  same  plane  and  there 
exists  but  one  plane  containing  one  of  the  parallel  lines 
and  a  point  on  the  other  (condition  1). 

EXERCISES 

1.  Illustrate  each  of  the  4  conditions  named  above  on  a 
cube. 

2.  Illustrate  the  same  facts  by  using  Hnes  and  points  in  the 
classroom. 

140.  Relative  positions  of  two  straight  lines.    Two 

straight  lines  in  space  may  have  the  following  relative 
positions : 

1.  They  may  intersect,  produced  if  necessary. 

2.  They  may  be  parallel. 

3.  They  may  not  he  parallel  and  not  intersect, 

EXERCISES 

1.  Illustrate  these  three  possibihties  by  selecting  the  proper 
edges  of  a  cube. 

2.  Find  other  illustrations  in  the  classroom. 

141.  Relative  positions  of  a  straight  line  and  a  plane. 
A  straight  line  and  a  plane  may  have  the  following  relative 
positions : 

1.  The  straight  line  may  intersect  the  plane,  produced 
if  necessary. 

2.  The  straight  line  may  be  parallel  to  the  plane,  i.e., 
have  no  point  in  common  with  the  plane,  however  far 
produced. 


QUADRILATERALS  55 

3.  The  straight  line  may  have  two  points  in  common 
with  the  plane  and  therefore  lie  entirely  within  the 
plane. 

Illustrate  each  of  these  cases  on  the  cube  and  on  lines 
and  planes  in  the  classroom. 

142.  Representation  of  a  plane  in  space.  A  plane  is 
conveniently  represented  by  a 

plane  figure,  such  as  a  rec- 
tangle, parallelogram,  etc., 
Fig.  56.     However,  the  figure  p      ~ 

indicates  only  the  position  of 
the  plane,  the  plane  itself  being  regarded  as  indefinite  in 
extent. 

143.  Theorem:  //  two  planes  inter  sect  j  the  intersection 
is  a  straight  line. 

Given  two  intersecting  planes 
P  and  Q,  Fig.  57. 

To  prove  that  P  and  Q  inter- 
sect in  a  straight  line.  *        "~Fiq~57 

Proof  (indirect  method)-: 

Suppose  the  intersection,  AB,  of  planes  P  and  Q 
not  to  be  a  straight  line. 

Then  it  will  be  possible  to  find  three  points  on  AB  not 
in  the  same  straight  line. 

Since  these  three  points  lie  on  the  intersection,  they 
must  be  in  both  planes,  P  and  Q. 

Therefore  P  and  Q  must  coincide.        Why  ? 

This  contradicts  the  hypothesis  in  which  P  and  Q  are 
understood  to  be  two  different  planes.  Hence,  the  assump- 
tion, that  the  intersection  of  P  and  Q  is  not  a  straight  line, 
is  wrong. 

Therefore  the  intersection  of  P  and  Q  is  a  straight  line. 


56 


SECOND-YEAR  MATHEMATICS 


Diedral  Angles 

144.  Diedral  angles.     Two  intersecting  planes  form 
a  diedral  angle,  Fig.  58.     The  planes  are  the  faces  and 
the   line   of  intersection   is  the 
edge  of  the  diedral  angle.    Point 
out  diedral  angles  in  the  class- 
room and  on  the  cube. 

A  diedral  angle  is  named  by 
two  points  in  the  edge  and  an 
additional    point   in   each   face. 

Thus,  the  diedral  angle  in  Fig.  58  is  denoted  C-AB-D. 
Sometimes  it  is  sufficient  to  name  only  two  points  on  the  edge, 
&sAB. 


Fig.  58 


145.  Size  of  diedral  angles.  A  diedral  angle  may  be 
formed  by  rotating  a  plane  about  a  line  in  the  plane.  The 
size  of  the  diedral  angle,  therefore,  depends  upon  the 
amount  of  rotation,  not  upon  the  extension  of  the  faces. 

146.  Plane. angle.  If  at  a  point  in  the  edge  of  a 
diedral  angle  two  lines  are  drawn  perpendicular  to  the 
edge,  one  in  each  face,  the  angle  formed  is  the  plane  angle 
of  the  diedral  angle. 

Thus,  ABC,  Fig.  59,  is  the  plane  angle 
of  P-QR-S. 

A  plane  angle  may  be  drawn  at  any 
point  of  the  edge. 

It  will  be  shown  in  §  380  that  all  plane 
angles  of  a  diedral  angle  are  equal. 

Fig.  59 

147.  Classification  of  diedral  angles.    A 

diedral  angle  is  said  to  be  right,  straight,  acute,  obtuse, 
reflex,  oblique,  according  as  the  plane  angle  is  right, 
straight,  etc.     Diedral  angles  are  adjacent  if  they  have 


QUADRILATERALS  57 

a   common   edge   and   a   common  face   between   them. 
Thus,  A-BC-D  and  D-BC-E, 
Fig.  60,  are  adjacent  diedral 
angles. 

Two  diedral  angles  are 
complementary  or  supple- 
mentary ^-ccording  as  the  plane 
angles  are  complementary  or  -p     ^p. 

supplementary. 

148.  Perpendicular  planes.     Two  planes  are  perpen- 
dicular to  each  other,  if  they  form  a  right  diedral  angle. 

Point  out  perpendicular  planes  on  the  cube;  in  the 
classroom. 

Summary 

149.  The   chapter   has  taught  the  meaning   of  the 


following  terms : 

parallelogram 

kite 

plane  angle  of  a  diedral 

rhomboid 

prismatic  surface 

angle 

rhombus 

prism 

perpendicular  planes 

rectangle 

base  of  prism 

analysis 

square 

lateral  surface 

converse  of  a  theorem 

trapezoid 

lateral  faces 

axis  of  symmetry 

isosceles  trapezoid  diedral  angle  locus 

150.  The  following  theorems  have  been  proved: 

1.  Two  parallelograms  are  congruent,  if  two  adjacent 
sides  and  the  included  angle  of  one  are  equal,  respectively, 
to  the  corresponding  parts  of  the  other. 

2.  A  parallelogram  may  he  constructed  if  two  adjacent 
sides  and  the  included  angle  are  given. 

3.  Two  lines  perpendicular  to  the  same  line  are  parallel. 

4.  If  two  alternate  interior  angles  formed  by  two  lines 
and  a  transversal  are  equal,  the  lines  are  parallel. 


58  SECOND-YEAR  MATHEMATICS 

5.  7/  two  parallel  lines  are  cut  by  a  transversal,  the  alter- 
nate interior  angles  are  equal, 

6.  //  a  quadrilateral  is  a  parallelogram — 

1.  A  diagonal  divides  it  into  congruent  triangles; 

2.  The  opposite  sides  are  equal; 

3.  The  opposite  angles  are  equal; 

4.  The  consecutive  angles  are  supplementary; 

5.  The  diagonals  bisect  each  other. 

7.  A  quadrilateral  is  a  parallelogram  if — 

1.  The  opposite  sides  are  parallel; 

2.  The  opposite  sides  are  equal; 

3.  One  pair  of  opposite  sides  are  equal  and  parallel; 

4.  The  opposite  angles  are  equal; 

5.  The  diagonals  bisect  each  other. 

8.  If  two  planes  intersect,  the  intersection  is  a  straight  line. 

151.  Quadrilaterals  have  been  classified  as  follows: 

(p      ,,  ,  f Rhomboid — Rhombus 

[Rectangle — Square 
Trapezoid — Isosceles  Trapezoid 

152.  The  following  methods  of  proof  have  been 
taught:  (1)  the  indirect  method,  (2)  the  method  of 
analysis. 

153.  Quadratic  equations  were  solved  by  factoring,  or 
by  completing  the  square. 

154.  Each  of  the  following  conditions  determines  the 
position  of  a  plane  in  space: 

1.  A  straight  line  and  a  point  not  in  that  line; 

2.  Three  points  not  in  the  same  straight  line; 

3.  Two  intersecting  straight  lines; 

4.  Two  parallel  straight  lines, 


CHAPTER  V 

PROPORTIONAL  LINE-SEGMENTS 

Uses  of  Proportional  Line-Segments 

155.  Measurement  of  line-segments.    To  measure  a 

•line-segment  is  to  find  how  many  times  it  contains  another 
Une-segment,  called  the  unit-segment.  The  number  of 
times  a  segment  b  is  contained  in  a  segment  a  is  the 
nimierical  measure  of  a  in  the  unit  b,  or  the  numerical 
measure  of  a  with  respect  to  b. 

156.  Ratio  of  two  segments.  The  ratio  of  the  numeri- 
cal measures  of  two  segments,  both  being  measured  with 
the  same  unit,  is  the  ratio  of  the  two  segments.  Another 
method  of  finding  the  ratio  of  two  segments  is  given  in 
§162. 

157.  Proportion.    An  equation  of  two  equal  ratios, 

4213ac.„,  ^.         ^ 

as  ^ =^,  c  =TE>  h~^}^  called  a  proportion.  Four  magni- 
tudes are  said  to  be  in  proportion,  if  their  numerical 
measures  are  proportional. 

Thus,  if  the  ratio  of  the  rectangles,  Fig.  61,  is  4  and 
if  the  ratio  of  the  altitudes  is  also  f,  the  rectangles  are 
proportional  to  the  altitudes. 


Fig.  61 


Fig.  62 


Show  that  the  central  angles,  AOB  and  A'O'B',  in 
Fig.  62,  are  proportional  to  the  intercepted  arcs. 

59 


60 


SECOND-YEAR  MATHEMATICS 


Fig.  63 


158.  Uses  of  proportional  line-segments 
represents  a  pair  of  proportional  compasses, 
used  to  make  scale  drawings  of  given  figures. 
By  making  OB' =  lOB  and  OA' =  lOA  and 
by  opening  the  compass  so  that  AS  equals 
a  given  line-segment,  we  obtain  A'B'  equals 
I A  5 .  This  fact  follows  from  one  of  the  prin- 
ciples of  proportional  line-segments  (§167). 

The  pantograph,  Fig.  64,  is  used  to  draw 
figures  to  definite  scales,  and  to  enlarge  or 
to  reduce  maps,  drawings,  designs,  etc.  The 
instrument  consists  of  four  pointed  bars, 
making  BBiA^A  a  parallelogram.  According  to  the  prin- 
ciples of  proportional  line-segments,  if  ~  is  made  equal 

B1A2 
to  -j^-^j  points  0,  A  J  and  Ai 

must  fall  in  a  straight  line, 

1  .      OA     OB      ^^      . 
making  ^^  =  ^^.    Keepmg 

point  0  fixed,  point  A  is  made 
to  describe  figure  (a).  The 
pencil  at  Ai  will  then  describe 
figure  (6),  which  is  figure  (a) 
magnified  to  the  scale  OBi  to  OB, 

The  diagonal  scale,  Fig.  65,  is  another  instrument 
whose  construction  is  based  upon  principles  of  propor- 


i"    .b"   .6" 

.V    .2" 

9 

/"                                           2" 

-    \-\     r 

/ 

8 

I-  ■  -l 

V 

L-i    I                    ■■- 

) 

6 

^1 

^       / 

1 

( 

A 

\ 

) 

'? 

11111 

( 

I 

1    1        1. 

\ 

Fia65 


PROPORTIONAL  LINE-SEGMENTS 


61 


tional  line-segments.     By  means  of  it  lengths  may  be 
measured  to  hundredths  of  an  inch. 


Fig.  66 


Fig.  66  represents  part  of  Fig.  65,  enlarged. 
BBi 


By  §  167, 
Hence, 
Similarly, 
and 

Since 

we  have 
Likewise, 


XXi 


AB 
AX 


1^ 
10 


BBi  =  ^XXi 
2 

CCi  =  jt:  XXi 

Z)Z)i=^ZZi,etc. 
1  1" 

BBi=  .01",  CCi=  .02",  DDi=  .03",  etc. 

5^2  =.1"+.  01"=.::" 

CC2=.1"+.02"=.12" 
Z)D2=.l"+.03"=.13",etc. 


62  SECOND-YEAR  MATHEMATICS 

EXERCISES 

1.  What* is  the  length  of  AB,  Fig.  65  ? 

2.  Draw  a  line-segment.     Measure  it  to  hundredths  of  an 
inch  by  using  the  diagonal  scale,  Fig.  65. 

Proportional  Segments 

159.  Theorem:*  A    line  that   bisects   one   side   of  a 
triangle,  and  is  parallel  to  a  second  side,  bisects  the  third 


Given   AABC,  CD  =  DA,  /f 

DE  II  AB,  Fig.  67. 

To  prove  CE  =  EB.  /  /^ 

Analysis:  How  may  two  line-      i  'f 

segments  be  proved  to  be  equal?  Fiq.  67 

Draw  EF  \\CA, 

CE  will  equal  EB,  if  ADEC^AFBE. 

Proof:  AFED  is  a  parallelogram.         Why? 

.\FE  =  AD.        Why? 
Show  that  FjE;  =  CD. 
Show  that     x  =  x\y  =  y'. 

.-.  ADEC  ^  AFBE.        Why  ? 

/.CE=EB.        Why? 

EXERCISES 

1.  Prove  that  CD,  DA,  CE,  and  EB,  Fig.  67,  are  propor- 
tional. 

2.  Prove  that  DE^^AB.    State  this  fact  in  form  of  a 
theorem. 

3.  If  (7Z)  =  3,  Z)^=3,  C^=4,  and  EB  =  x-2,  find  x  and  the 
length  of  EB. 

*This  theorem  is  probably  to  be  credited  to  Eudoxus  (408- 
355  B.C.). 


PROPORTIONAL  LINE-SEGMENTS 
3a;- 1    4x-5 


63 


4.  If  CD  =  1,  Z)A  =  1,   CE==     ^ 
find  X  and  the  lengths  of  CE  and  EB. 


.EsJ-^-i, 


160.  Theorem;*  If  three  or  more  parallel  lines  inter- 
cept equal  segments  on  one  transversal  they  intercept  equal 
segments  on  every  transversal. 


Proof  (method  of  congru- 
ent triangles) :  Draw  helping 
lines  a"  II  a,  h"  II  6,  etc., 
Fig.  68. 

Prove  AI^  All  ^AIII, 
etc. 

Then      a'  =  6'  =  c',  etc. 


a  I  i\a ' 


^/iV' 


/iiK^c' 


d'lv^ 


Fig.  68 


Why? 


j^' 


\ 


EXERCISES 

1.  Prove  segments  a,h,c, ^a\h\c\ ,  Fig.  68, 

proportional. 

2.  A  line   drawn    through    the 
midpoint  of  one  of  the  non-parallel 
sides  of  a  trapezoid,  parallel  to  the 
bases,  bisects  the  other  side.    Prove.     ^ 
Apply  §  160. 


Fig.  69 


161.  Median  of  a  trapezoid.  The  segment  joining 
the  midpoints  of  the  non-parallel  sides  of  a  trapezoid  is 
the  median  of  the  trapezoid. 


EXERCISE 


Prove  that  the  median  of  a  trapezoid 
equals  one-half  the  sum  of  the  bases. 

Show  that  mi  =  |6i,  m2  =  ^b2,  Fig.  70; 
.'.  w=wi +7712  =  1(61+62). 


A.  .<\ 

/'-         \ 

Fig.  70 

*  This  theorem  is  attributed  to  Archimedes  (287-212  B.C.). 


64 


SECOND-YEAR  MATHEMATICS 


162.  Ratio  of  line-segments.  The  ratio  of  two  line- 
segments  may  be  found  by  means  of  the  compass  as 
follows : 

Let  AB  and  CD,  Fig.  71,  be  two  segments  whose  ratio  is  to  be 
found, 

EG 

27  27  27  2  21- 

CI   ,    I   ,   I   ,    I   ,   I       ^IP 
5       5       5       5       5     2 

Fig.  71 

Let  us  assume  that  AB  and  CD  contain  a  common  unit  of 
measure.  It  will  be  shown  in  §  165  that  there  are  Une-segments 
that  have  no  common  unit  of  measure.  To  find  the  common 
unit,  proceed  as  follows: 

Lay  off  the  smaller  segment  CD  on  the  larger  AB  as  often  as 
possible,  leaving  a  remainder  EB,  which  is  less  than  CD. 

Lay  off  EB  on  CD,  leaving  a  remainder  FD,  which  is  less 
than  EB. 

Lay  off  FD  on  EB,  leaving  a  remainder  GB. 

Lay  off  GB  on  FD,  leaving  no  remainder. 

The  last  remainder,  GB,  is  a  common  unit  of  measure  of  AB 
and  CD. 

Using  GB  as  unit,  show  that  AB  =  86,  and  CD  =  27. 

Therefore  7n^  =  27 - 

163.  Theorem:  If  two  parallels  cut  two  intersecting 
transversals  the  segments  intercepted  on  one  transversal  are 
proportional  to  the  corresponding 

segments  on  the  other.  ct. 

Given    ABWDE    and    AD 

intersecting  BE  at  C,  Fig.  72,  dT             ^^ 

To  prove  

CD^CE    CD^CE^    M=^       ^^ ^ 

DA     EB '  CA     CB'  CA     CB  *  Fia.  72 


PROPORTIONAL  LINE-SEGMENTS  65 

CD 
Proof:    To  find  the  ratio  j^  j  ^^y  ^^  ^^^  smaller 

segment,  DA,  on  the  larger,  CD,  as  often  as  possible.  If 
there  is  a  remainder,  lay  it  off  on  AD.  If  there  is  still 
a  remainder  lay  it  off  on  the  preceding  remainder,  etc. 
We  will  assume  that  after  laying  off  these  remainders  a 
definite  number  of  times  there  is  no  remainder.  Then 
the  last  remainder  is  a  common  unit  of  CD  and  DA . 

Let  this  common  unit  be  contained  in  CD  and  DA, 
m  and  n  times,  respectively. 

^,       CD     m       1.     ^.      -^   CD    6 

To  find  the  value  of  ^^^  proceed  as  follows: 

Draw  lines  parallel  to  A  5  passing  through  the  points 
of  division  of  CA, 

These  lines  will  divide  CE  and  EB  into  m  and  n  parts, 
respectively.     Why  ? 

Moreover,  these  parts  are  equal  to  each  other.     Why  ? 

Hence,  g  =  ^        Why? 


CD  JCE 
DA~EB 


Why? 


cj.    .,    ,  CD    CE    DA    EB 

Similarly,  we  may  prove  ^  =  ^ ,  CA^CB 


EXERCISES 

1.  Prove  that  a  line  parallel  to  one  side  of 
a  triangle  divides  the  other  two  sides  propor- 
tionally.    (Apply  §  163.) 

t2.  Prove  the  theorem*  in  §  163,  using 
Fig.  73. 

*  This  form  of  the  theorem  is  attributed  to  Archimedes. 


66  SECOND-YEAR  MATHEMATICS 

164.  Commensurable  magnitudes.  In  the  proof  of 
the  theorem  of  §  163  it  was  assumed  that  a  common  unit 
for  CD  and  DA  could  be  found.  Two  magnitudes  which 
have  a  common  unit  of  measure  are  said  to  be  com- 
mensurable. 

165.  Incommensurable  magnitudes.  Not  all  mag- 
nitudes have  a  common  unit  of  measure.  Magnitudes 
not  having  a  common  unit  of  measure  are  said  to  be 
incommensurable. 

EXAMPLE 

The  side  and  diagonal  of  a  square  are  incommensurable 
segments.    This  may  be  seen  as  follows: 

Since  AB  <AC,  Fig.  74,  AB  may 
be  laid  off  on  AC,  leaving  a  remainder 
BxC. 

Draw  BiA  1I.AC  oXBu  Prove,  by 
congruent  triangles,  that  BiAi=AiB. 

FroyeBiC=BiAi  =  AiB. 

Since  BiCkAiC,  it  follows  that 
BiC  may  be  laid  off  on  AiC  leaving  a 
remainder,  as  B2C. 

Thus,  BiC  may  be  laid  off  on  BC  Fig.  74 

twice,  leaving  a  remainder  B2C. 

In  the  same  way  it  may  be  shown  that  B2C  may  be  laid  off 
on  BiC  twice,  leaving  a  remainder,  B3C;  that  B3C  may  be  laid 
off  twice  on  B2C,  leaving  a  remainder,  etc. 

In  each  case  the  process  is  a  repetition  of  the  preceding  case, 
only  with  smaller  segments.  Since  in  each  case  there  is  a 
remainder,  the  process  may  be  kept  up  indefinitely. 

Hence,  no  common  unit  of  AB  and  AC  can  be  found. 

AC        - 
It  will  be  seen  later,  §  254,  that  the  ratio  t»=  ^2,  is  an 

irrational  number,  i.e.,  a  number  which  cannot  be  expressed 
exactly  in  terms  of  integers,  or  of  fractions  whose  terms  are 
integers. 


PROPORTIONAL  LINE-SEGMENTS 


67 


X  166.  The  incommensxirable  case  of  the  theorem 
m  §  163.  Since  several  theorems  that  have  not 
yet  been  proved  are 
needed  for  proof  of  the 
incommensurable  case, 
only  an  outline  of  the 
proof  will  be  given. 
The  method  of  proof  is 
indirect. 


Assume 


Then  either 


EB 


FiQ.  75 


^^^,  Fig.  75. 


Let 


DA     EB 
CD^CE' 
DA    EB 
CD^CE' 


DA    EB 
^^  CD^CE 


Select  a  point  F  on  the  extension  of  EB,  making  EF 
.                 ,^.DAEF  ,,, 

long  enough  to  give  ^  =  ^ (1) 

It  is  possible  to  determine  a  point  H  between  B  and 

F,  making  CE  and  EH  commensurable. 

HK  II  BA 

DK^EH 

CD     CE 

DA<DK, 


Draw 
Then, 
Since, 
it  follows  that 


DA    EH 
CD^CE 


.(2) 


EF    EH 
Comparing  (2)  and  (1),  we  have  Trh<TrF> 

.'.    EF<EH 

This  is  impossible  and  the  assumption  that  777^  >  77^, 


IS  wrong. 


T)  A  FR 

Similarly,  we  may  prove  that  ^^  is  not  less  than  j^ 


Hence, 


DA 
CD 


EB 
CE' 


68 


SECOND-YEAR  MATHEMATICS 


167.  Theorem:*  If  a  nu7nber  of  parallel  lines  are  cut 
by  two  transversals,  the  segments  of  one  transversal  are  -pro- 
portional to  the  corresponding  segments  of  the  other. 


a    a 


Show  that    g=p(§163) 

Draw        DE  \\  AB 

Prove  that  -77=-? 
c"    c' 


But 
Why? 


h  =  h"  and  c  =  c' 


-7 ,  etc. 
c 


Fig.  76 


EXERCISE 

Prove  that  if  two  given  lines  are  cut  by  two  parallel  lines  the 
segments  of  the  parallel  lines  are  proportional  to  the  corresponding 
segments  of  the  given  lines. 

CD^DE 

CA    AB 

DF  II  CB. 


We  are  to  prove 
Draw 


Then, 


CD 
CA 


BF 
BA 


Fig.  77. 

DE 
'AB' 


Why? 


Fig.  77 


168.  Theorem:  Two  lines  that  cut  two  given  intersecting 
lines,  and  make  the  corresponding  segments  of  the  given  lines 
proportional,  are  parallel  (converse 
of  §  163). 

CD     CE    ^.     ^^ 
Fig.  78, 


Given 


EB 
DE 


A*— = 


DA 
To  prove  AB 

Proof  (indirect  method) : 

Suppose  AB  not  parallel   to   DE. 

Draw        AF  II  DE 


Fic.  78 


*  This  theorem  was  first  proved  by  Archimedes. 


PROPORTIONAL  LINE-SEGMENTS 


69 


Then, 


But, 


CD^CE 

DA~EF 

CD^CE 

DA~EB 

.  CE^CE 

"  EF    EB 

/.CE'EB  =  CE  'EF 

:.EB  =  EF.      Why? 
This  is  impossible.    Why  ? 

Therefore  the  assumption  that  AB  is  not  parallel  to 
DE  is  wrong  and  AB  II  BE. 


Why? 

Why? 

Why? 

Why? 


EXERCISES 


Prove  the  following: 


|1.  If  ^=^,  Fig.  78,  then  DE  \\  AB, 


2.  The  line  joining  the  midpoints  of  two  sides  of  a  triangle 
is  parallel  to  the  third  side. 

Prove  that  two  sides  are  divided  proportionally.     Then  apply 
§  168. 


JS.  Prove  §  168,  using  Fig.  79. 


3C      X 

4.  In  Fig.  80  a\]b  and    =— ,.     Prove 

y   y 

that  c  II  6. 


5.  The  median  of  a  trapezoid  is  parallel 
to  the  bases. 


^A 


6.  The    quadrilateral   whose    vertices  / ^ _\ 

are  the  midpoints  of  the  sides  of  a  triangle  y 

and  one  vertex  of  the  triangle  is  a  paral-  /             ^ 

lelogram.  Fig.  80 


70 


SECOND-YEAR  MATHEMATICS 


7.  The  midpoints  of  the  sides  of  a  quad- 
rilateral, Fig.  81,  may  be  taken  as  vertices  of 
a  parallelogram. 

Draw  the  diagonal.    Use  Exercise  2. 


Fig.  81 


Fig.  82 


/ 


Fig.  83 


Fig.  84 


169.  Summary  of  the  more  important  theorems  in 
proportional  segments  in  §§  159-68: 

1.  A  line  bisecting  one  side 
of  a  triangle  and  parallel  to  a 
second  side,  bisects  the  third 
side,  Fig.  82. 

2.  If  the  segments  inter- 
cepted by  parallel  lines  on  one 
transversal  are  equal,  then  the 
segments  intercepted  on  every 
transversal  are  equal.  Fig.  83. 

3.  The  line  drawn  through 
the  midpoint  of  one  of  the  non- 
parallel  sides  of  a  trapezoid 
parallel  to  the  bases,  bisects 
the  other  side.  Fig.  84. 

4.  A  line  parallel  to  one 
side  of  a  triangle  divides  the 
other  two  sides  proportionally, 
Fig.  85. 

5.  If  a  number  of  parallels 
cut  two  transversals  the  seg- 
ments of  one  transversal  are 
proportional  to  the  correspond- 
ing segments  of  the  other. 
Fig.  86. 

6.  A  line  parallel  to  the  base  of  a  trapezoid  divides 
the  two  non-parallel  sides  proportionally,  Fig.  87. 


Fig.  85 


II 


Fig.  86 


Fig.  87 


PROPORTIONAL  LINE-SEGMENTS 


71 


7.  Two  lines  that  cut 
two  intersecting  lines 
making  the  corresponding 
segments  proportional  are 
parallel,  Fig.  88. 

8.  The  line  joining  the 
midpoints  of  two  sides  of 
a  triangle  is  parallel  to  the 
third  side,  Fig.  89. 

9.  The  median  of  a 
trapezoid  is  parallel  to  the 
bases,  Fig.  90. 


Fig.  88 


III 


Fig.  89 


Fig.  90 


EXERCISES 


t  1.  If  the  segment  joining  the  midpoints  of  two  opposite 
sides  of  a  quadrilateral  and  a  diagonal  bisect  each  other,  the 
quadrilateral  is  a  parallelogram. 


2.  Prove  that  the  medians  and 
diagonals  of  a  parallelogram  meet  in 
a  common  point  (Fig.  91). 


Fig.  91 


170.  Theorem:   The  bisector  of  an  interior  angle  of  a 
triangle  divides  the  opposite  side 
into  segments  that  are  propor- 
tional to  the  adjacent  sides. 


Given   A  ABC,  x=y, 
Fig.  92, 

ABAC 
DB    CB 


To  prove 


72 


SECOND-YEAR  MATHEMATICS 


Proof:  Extend  BC 


Draw       AE  \\  DC 

^,            AD    EC 
Then,      ^^  =  ^^^ 

x=x' 

y=y' 

Why? 

Why? 
Why? 
Why? 
Why? 
Why? 

Why? 

E 

x  —  y 

/.    x'=y' 

.'.    EC  =  CA 

.     AD    AC 

"    DB    CB 

FiQ.  92 

EXERCISES 

1.  Prove  that  a  line  passing  through  the  vertex  of  a  triangle 
and  dividing  the  opposite  side  into  segments  proportional  to  the 
other  two  sides,  bisects  the  angle  included  between  those  sides 
(converse  of  §  170). 

To  prove  this  exercise,  use  the  proof  in  §  170  as  an  analysis. 

2.  In  Fig.  92,  AC  =  8,  (75=10,  AB  =  9.  Find  the  lengths 
of  AD  and  DB. 

3.  In  Fig.  92,  AC=5,  CB  =  4:,  DB  =  S.  Find  the  lengths  of 
AD  and  AB. 

4.  If  CA  =  8,CB=  16;  and  AB  =  12,  Fig.  92,  find  AD  and  DB. 

171.  External  division  of  a  segment.  A  point  P  on  a 
segment  AB  divides  AB  into  the  segments  ^P  and  PB, 
Fig.  93.  Considering  the 
direction  ^J5  as  positive, 
and  the  direction  BA  as 
negative,  then 
(+AP)  +  {+PB)  =  {+AB).  p^^^4 

If  P  is  on  the  extension 
of  AB,  Fig.  94,  then  AP  is  positive,  and  PP  is  negative. 


/i. 

P 

Fig.  93 

A 

B 

P 

PROPORTIONAL  LINE-SEGMENTS  73 

nevertheless  the  statement  {AP)-\-{PB)  ={+AB)  still 
holds  good.  Because  of  this  equation,  AP  and  PB  are 
called  parts  of  AB,  and  AB  is  said  to  be  divided 
externally  by  P.  Thus,  in  external  as  in  internal  division 
oi  AB  the  two  parts  are  measured  one  from  A  to  P,  and 
the  other  from  P  to  B. 

172.  Theorem:  The  bisector  of  an  exterior  angle  of  a 
triangle  divides  the  opposite  side  externally  into  segments 
that  are  proportional  to  the  other  sides.  * 

Given  AABC,x  =  y. 

AD    AC 
To  prove    ;^=^. 

The  proof  is  practi-  ^^  "N^ 

cally  the  same  as  in  §  170.  ^.  -""     /    ^^.^^ 

173.  Harmonic  divi-  ^^-""  fe-"'""  ^\s. 
sion.     If   a   segment   is  ^^                       "a 

divided    internally    and  ^iQ-  95 

externally    in    the    same 

ratio  it  is  said  to  be  divided  harmonically. 

EXERCISE 

Prove  that  the  bisector  of  an  interior  angle  of  a  triangle 
and  the  bisector  of  the  exterior  angle  at  the  same  vertex  divide 
the  opposite  side  harmonically. 


Problems  of  Construction 

a     c 
174.  Fourth  proportional.     In  a  proportion,  asT=;7, 

d  is  the  fourth  proportional  to  a,  h,  and  c. 

*  Pappus  of  Alexandria  recognized  this  theorem,  though  the 
Pythagoreans  were  the  first  to  deal  with  the  harmonic  division  of 
lines  (Tropfke,  History  of  Elementary  Mathematics  [in  German], 
Vol.  II,  p.  82). 


74  SECOND-YEAR  MATHEMATICS 

EXERCISES 

1.  To  construct  the  fourth  proportional  to  three  given  segments. 

Given  the  segments  a,  h,  and  c. 

Required  to  construct  the  fourth  proportional  to  a,  6,  and  c. 

a)  Algebraic  solution:  Let  x  be  the  fourth  proportional. 
Find  the  values  of  a,  b,  and  c  by  measuring  and  substitute  them 

in  the  proportion  i  =  ~  • 

Solve  this  equation  for  x. 

Construct  a  segment  whose  measure  is  x. 

This  is  the  required  fourth  proportional. 

b)  Geometric  solution :  On  one  of  two  intersecting  lines,  as 
AB,    lay    off    AD  =  a,    DE=b, 

Fig.  96. 

On  the  other,  as  AC,  lay  off 
AF=c. 

Draw  DF. 

Draw  EG  \\  DF, 

Then  FG  is  the  required 
fourth  proportional.  Prove  by 
§163. 

To  test  the  correctness  of  the  construction,  measure  the  four 
segments  to  two  decimal  places  and  see  if  these  four  numbers  are 
proportional. 

2.  Find  by  means  of  an  equation  the  fourth  proportional  to 
1,  2,  and  8. 

X      4 

3.  Solve  for  x:     h,  =  T^» 

57     16 

175.  Third  proportional.  In  a  proportion,  as  7=-, 
c  is  called  the  third  proportional  to  a  and  6. 

EXERCISE 

Construct  the  third  proportional  to  two  segments:  (a)  alge- 
braically; follow  the  instructions  of  (a),  exercise  1,  §  174; 
(6)  geometrically,  as  in  (6),  §  174, 


h                             K. 

^B 

c                      h  ^ 

D^         N 

\ 

a^^  \ 

\ 

\ 

'^             0         F 

G        • 

Fig.  96 

PHOTOGRAPH  OP  500-DRAW  SPAN,  SHOWING  CHANNEL  OPEN 


CONSTRUCTION  OF   RAILWAY   BRIDGE    IN    SIERRA    LEONE, 
WEST  AFRICA 

Point  out  the  uses  of  mathematical  forms  in  bridge  and  trestle 
construction,  using  the  structures  shown  above  as  illustrations. 


PROPORTIONAL  LINE-SEGMENTS  75 

176.  To  divide  a  segment  in  a  given  ratio.    To  divide 

177, 

a  segment  AB  internally  in  the  ratio  —  means  to  find' a 

point,  P,  on  A B  so  that  p^  =  — .    To  divide  AB  externally 

in  the  ratio  —  means  to  find  a  point,  P\  on  the  extension 
n 

of  AB  so  that  y^  =  —  (see  §  171). 
P  B     n 


EXERCISES 

772 

1.  To  divide  a  segment  internally  in  the  ratio  — 

Let  AB  (Fig.  97)  be  the  given 
segment.  Draw  a  line  AC  through  A 
and  lay  off  AD  =  m  and  DE  =  n.  Draw 
EB.    Through  D  draw  DF  \\  EB.    Then 

F  divides  AB  internally  in  the  ratio  —  . 

Test  the  correctness  of  the  construc- 
tion by  measuring  the  segments.     Give  ■^^^-  ^* 
proof. 

2.  Show  how  to  divide  a  segment  internally  in  the  ratio  — , 
using  §  170. 

m 

3.  To  divide  a  segment  externally  in  the  ratio  — . 

DrawAD  =  w(Fig.98),DE=n.    Join^to5anddrawDi?"l|  EB, 

^,       AF'    m      ^ 
Then  ^^j^r  =—  .    Prove. 
F'B     n 


m 

n 

p 

,-'• 

P1^' 

,'< 

\ 

-i^.' 

''    \ 

\ 

,y' 

\ 

\ 

A-^ 

: 

— ^R 

F 


Fig.  98 


4.  Show  how  to  divide  a  segment  externally,  using  §  172. 


76 


SECOND-YEAR  MATHEMATICS 


5.  A  segment  AB  =  18  is  divided  internally,  or  externally,  at 

AP 
a  point  P.    What  is  the  ratio  p^  for  AP  =  2?    3?    6?    9? 

20?     30? 


6.  To  divide  a  given  segment,  AB,  into  segments  proportional 
to  several  given  segments,  x,  y,  and  z. 

On  a  line,  as  AC  (Fig.  99),  lay  off  x,  y,  z  sliccessively  and  join 
B  to  the  last  point  of  division  D.    Draw  parallels  to  BD  at  the 

points  of  division.    Then  — ,  =  - 
y    y 


z'     z' 


Why? 
Why? 


:N.a: 


-v> 


D 


&         B 


,->v 


,^V. 


Fig.  99 


'-^D 


Fig.  100 


7.  To  divide  a  segment  into  equal  parts  (Fig.  100). 

The  construction  is  the  same  as  in  exercise  6,  using  equal 
segments  instead  of  x,  y,  and  z. 


Lines  and  Planes  in  Space 

177.  Line  perpendicular  to  a  plane.  If  a  straight  line 
intersects  a  plane  and  is  perpendicular  to  every  straight 
line  passing  through  the  point  of  intersection  and  lying 
in  the  plane,  it  is  said  to  be  perpendicular  to  the  plane. 

Show  that  the  vertical  edge  of  a  door  is  perpendicular  to 
the  floor  of  the  classroom. 

Show  that  an  edge  of  a  cube  is  perpendicular  to  one  of  the 
faces. 


PROPORTIONAL  LINE-SEGMENTS 


77 


178.  Theorem:    Two  planes  perpendicular  to  the  same 
line  are  parallel. 

Given   planes    P  and  Q  per- 
pendicular to  AB.  y 

To  prove     P  II  Q.*  ^ 


Proof  (indirect  method) : 
Suppose  P  is  not  parallel  to  Q. 
Then  P,  if  far  enough  extended, 
meets  Q  in  some  point,  C. 

Imagine  CA  and  CB  drawn. 
Then  CA  lies  wholly  in  plane  P. 
C5  Hes  in  plane  Q.        Why? 


F 


-y 


Fig.  101 


Why? 

.'.  CA  and  CB  are  both  perpendicular  to  AB 
(§  177).   ^ 

This  is  impossible,  as  only  one  perpendicular  can  be 
drawn  from  a  point  to  a  line. 

Therefore,  the  assumption  is  wrong  and  P  is  parallel 
toQ. 

179.  Theorem;    //  two  parallel  planes  are  cut  by  a 
third  plane,  the  intersections  are 
parallel. 

Given    plane    P  II  plane    Q, 

plane  R  intersecting  planes  P  and 

Q  in  AB  and  CD  respectively; 

Fig.  102. 

To  prove    AB\\  CD. 

Fig.  102 

*  When  proving  theorems  involving  lines  and  planes  in  space, 
the  student  will  find  it  helpful  to  think  of  Unes  and  planes  in  the 
classroom  as  representing  the  conditions  of  the  theorem.  Thus, 
the  ceiling,  the  floor  and  the  line  of  intersection  of  two  walls  will 
illustrate  the  conditions  of  this  theorem. 


78 


SECOND-YEAR  MATHEMATICS 


Proof  (indirect  method) : 
Assume  AB  not  parallel  to  CD. 
Since  AB  and  CD  lie  in  plane 

R  they   would  meet,    if  far 

enough    extended,    at    some 

point  E. 
Then  E,   being  on  both  lines 

AB  and  CD  would  lie  in  both 

planes  P  and  Q.         Why? 
This  contradicts  the  hypothesis  that  P  ||  Q, 
Therefore  the  assumption  is  wrong  and  AB 


Fig.  102 


CD 


180.  Theorem:    Parallel  line-segments  intercepted  by 
parallel  planes  are  equal. 


Prove  ACDB,  Fig.  103,  a  parallelo- 
gram. 

Then  AB  =  CD.        Why? 


Fig.  103 


181.  Theorem:  If  three  or  more  par- 
allel planes  are  cut  by  two  transversals,  the 
corresponding  segments  of  the  transversals  are  in  proportion. 

Given  planes  P  II  Q  II  R, 

cut  hyAB  and  CD,  Fig.  104. 

AE    CF 
To  prove    ^=|^. 


Proof:  Draw  CB, 
cutting  Q  in  K.  Pass 
planes  through  AB  and 
BC,  and  BC  and  CD,  . 
cutting  planes  P,  Q,  and  R 
in  AC,  EK,  KF,  and  BD. 
PWQ 
.-.     AC  II  EK 


/r-- 

-^C 

7 

/       1         /K—- 

9^F 

7 

^^ 

^ 

/.ft — - 

,      R. 

Fig.  104 


Why? 
Why? 


PROPORTIONAL  LINE-SEGMENTS  79 

QWR  Why? 

.-.      KFWBD       Why? 


AE^CK 
EB    KB 


Why? 


^^^  m  =  %        '''^^^' 


AE^CF 
EB    FD 


Why 


? 


Summary 

182.  The  chapter  has  taught  the  meaning  of  the  fol- 
lowing terms: 

proportion  internal  and  external  division 

diagonal    scale,    pantograph,  of    a    segment,    harmonic 

proportional  compasses  division 
median  of  a  trapezoid  fourth  proportional,  third  pro- 
commensurable    and   incom-  portional 

mensurable  magnitudes  line  perpendicular  to  a  plane 

183.  The  following  theorems  have  been  proved : 

1.  A  line  bisecting  a  side  of  a  triangle  and  parallel  to  a 
second  side  bisects  the  third  side. 

2.  If  three  or  more  parallel  lines  intercept  equal  segments 
on  one  transversal,  they  intercept  equal  segments  on  every 
transversal. 

3.  If  two  parallels  cut  two  intersecting  transversals, 
the  segments  intercepted  on  one  transversal  are  proportional 
to  the  corresponding  segments  on  the  other. 

4.  If  a  number  of  parallels  cut  two  transversals  the 
segments  intercepted  on  one  transversal  are  proportional 
to  the  corresponding  segments  on  the  other. 


80  SECOND-YEAR  MATHEMATICS 

5.  Two  lines  that  cut  two  given  intersecting  lines  and 
make  the  corresponding  segments  of  the  given  lines  propor- 
tional, are  parallel. 

6.  The  line  joining  the  midpoints  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side. 

7.  The  bisector  of  an  interior  (exterior)  angle  of  a  triangle 
divides  the  opposite  side  internally  (externally)  into  seg- 
ments that  are  proportional  to  the  adjacent  sides. 

8.  Two  planes  perpendicular  to  the  same  line  are  parallel. 

9.  //  two  parallel  planes  are  cut  by  a  third  plane  the 
intersections  are  parallel. 

10.  Parallel  segments  intercepted  by  parallel  planes  are 
equal. 

11.  If  three  or  more  parallel  planes  are  cut  by  two 
transversals,  the  corresponding  segments  of  the  transversals 
are  in  proportion. 

184.  The  following  constructions  have  been  taught: 

1.  To  construct  the  fourth  proportional  to  three  given 
line-segments. 

2.  To  construct  the  third  proportional  to  two  segments. 

3.  To  divide  a  segment  in  a  given  ratio,  internally  and 
externally. 

4.  To  divide  a  segment  into  parts  proportional  to  several 
given  segments. 

5.  To  divide  a  segment  into  equal  parts. 


CHAPTER  VI 

PROPORTION.    FACTORING.    VARIATION 

Fundamental  Theorems 

185.  In  the  first-year  course  we  saw  the  importance 
of  proportions  in  the  solution  of  problems.  In  chapter  v 
we  made  a  study  of  proportional  line-segments.  It  is 
one  of  the  purposes  of  this  chapter  to  study  the  properties 
of  proportions. 

186.  Theorem:  In  a  proportion  the  product  of  the 
means  is  equal  to  the  product  of  the  extremes. 

Proof:     Multiply    both    members    of    the    equation 

hi  ^y  *'^- 

The  preceding  theorem  is  important  because  it  is  a 
convenient  test  of  proportionality,  and  also  because  it 
suggests  a  simple  way  of  clearing  of  fractions  such  equa- 
tions as  are  proportions. 

EXERCISES 
Using  the  theorem  in  §  186,  work  the  following  exercises: 

1.  Which  of  these  statements  are  proportions  ? 

9~6'    15~7'\i8~21'^~20* 
81        ^^^-^-"^^ 


82  SECOND-YEAR  MATHEMATICS 

2^  Clear  the  following  equations  of  fractions,  but  do  not 
solve  them: 

4^20      180-a;_  5       2±x_5±x 

x-y__         7  _8 u^-uv-j-v^ 

10      x^-\-xy-{-y^ '    u+v~        2         ' 

?+?=^±?.         a;-l x+2 

x-4:    x-7'    x^-^Sx-\-4:~2x^-4x+7' 

X     12 
3.  Solve  the  equation  o  = "«  • 

187.  The  following  exercises  show  that  proportions 
may  be  obtained  from  equations  that  express  equality 
of  products. 


EXERCISES 

1.  The  statements  below  are  different  arrangements  of  the 
four  factors  in  the  equation  8  •  7  =  14  •  4.  Some  of  them  are 
equations,  others  only  appear  to  be  equations.  Apply  the 
test  of  proportionaliti/  and  point  out  which  statements  are 
proportions. 


1  ^-* 

^-  14~7 

3      ^--^ 
'•     7-14 

B     8-14 

7.     ?  =  ^ 
4     14 

2    TJA 

■'•4     8 

4    li_Z 
*•    8  -4 

6    ^       8 
^•14-4 

8    ^-^ 
^-  14-4 

2.  Exercise  1  shows  that  proportions  are  formed  from  the 
numbers  4,  7,  8,  and  14  only  when  they  are  taken  in  a  certain 
order.  From  what  place  in  the  equation  8  •  7  =  14  •  4  must  the 
means  be  taken  to  form  a  proportion  ?    The  extremes  ? 

3.  Write  four  proportions  from  3  •  28  =  4  •  21.  Apply  in 
each  case  the  test  of  proportionaHty. 


PROPORTION.     FACTORING.    VARIATION  83 

4.  Write  four  proportions  from  a  •  126  =  3a  •  46,  and  test. 
Exercises  1  to  4  illustrate  the  following  theorem : 

188.  Theorem:  //  the  product  of  two  factors  is  equal 
to  the  product  of  two  others,  proportions  may  be  formed  by 
taking  as  means  the  factors  of  either  product,  and  as 
extremes  the  factors  of  the  other  product. 

Given  ad  =  bc. 

a     c 
To  prove  that    t=-t 
0     a 

Proof:  Divide  both  members  of  the  equation  ad  =  bc 
by  bd. 

EXERCISES 

1.  Let  ad=bc.  Prove  that  the  following  statements  are 
proportions: 

a_b  b  __d  c     d  c     a 

c     d  a    c  a~b  d~b 

2.  Form  proportions  from: 

1.  5a-10b  =  ^^-3xy 

2.  lQa^^2axy  =  ax-\-ay—az 

3.  ix+y)^=m^-2mx+x^ 

4.  i^2+8Z+16  =  62-66+9 

5.  16a2-2562  =  36-25?/2 

6.  a2-62  =  c2-d2 

7.  p4-16  =  a2-64 

8.  ax-\-ay-\-az  =  br-\-bs-{-bt 

9.  5m2+10mn-15n2=9a2-4a6-1362 

10.  Qx^+lSx+2=a^-{-2a-\-l 

11.  a;2-5x+6  =  2/2+3i/-28 


84  SECOND-YEAR  MATHEMATICS 

Factoring 

189.  Review:  In  arithmetic  we  have  tests  of  divisi- 
biUty  by  which  we  can  tell  when  2,  3,  5,  9,  11,  etc.,  are 
divisors  of  a  number.  Likewise,  in  the  course  of  the  first 
year  we  have  learned  how  to  recognize  factors  of 
certain  polynomials.  This  work  may  be  summarized  as 
follows : 

Poljniomials:  Common  monomial  factor,  as  ax-\-ay. 
In  this  case  the  common  factor  is  one  of  the  factors  of 
the  polynomial.  The  other  factor  is  found  by  dividing 
the  polynomial  by  the  common  factor. 

Thus,  ax^-ay  =  a{x^-y). 

Factor  the  following : 

1.  3a:+32/  6.  8a;Y+4xy 

2.  ca;2+dx3+/a;4  6.  ^xY-'^^y-^^V^ 

3.  ba^h+24.a'^c-lOaH  7.  l^aH-lWy+ba^z 

4.  3a26-12a62  8.  ^2aW-ab^ 

Binomials:    The  difference  of  two  squares,  as  x^—y^. 
The  factors  are :   the  difference  of  the  square  roots  of 
these  squares  and  the  sum  of  the  square  roots. 


Thus,                Jc2- 

-y'=ix- 

-y)(x+y). 

Factor  the  following: 

1.  l-144a;V 

6.  9a2-16 

2.  ^:-i 

y' 

7.  16a2-25 

8.  1-7^ 

3.  x^-25y^ 

9.  4rc2-9?/2 

4.  (6+c)2-a2 

10.  a^-6^ 

5.  {a+hy-ic+dy 

11.  x^-y^ 

PROPORTION.    FACTORING.    VARIATION 


85 


Trinomials:  (1)  Trinomial  Squares,  as  x^-^2xy-\-y^  and 
x'^—2xy-^y^. 

In  each  case  we  have  two  equal  factors,  i.e.,  the  sum 
of  the  square  roots  of  the  square  terms  if  the  sign  of  the 
remaining  term  is  +,  and  the  difference  of  the  square 
roots  of  the  square  terms  if  the  sign  of  the  remaining 
term  is  — . 

Thus,  x'-{-2xy-\-y'=ix+yy 
and     Jc2-2xy+y2=  (x-y)2. 


Factor  the  following: 

1.  4m^  —  12am-\-9a^ 

2.  a2-8a+16 

3.  9+30x+25x2 

4.  3Qx^+25y^-Q0xy 


5.  25+80r+64r2 

6.  c2-16c+64 

7.  x^+30a:2+225 

8.  121a2+198a2/+8l2/2 


(2)  Trinomials  of  the  form  ax'^-^hx-\-c.     The  factors 
are  found  by  trial. 

Thus,  for  factors  of  3a;2+17x+10  we  have  as  one  of 

{Sx-^2 
_.     Multiplying,  we  find 
x-\-o 

that  Sx^+17x+10={Sx+2)(x+5). 

Factor  the  following: 
1.  2a;2+lla;+12 


2.  8c2+46c-1262 

3.  3a;2-17x+10 

4.  lla2-23a6+262 


5.  8?/2-31i/+21 

6.  5x2-380;+ 21 

7.  7/b2+123/c-54 

8.  5w2-29mn+36n2 


86  SECOND-YEAR  MATHEMATICS 

190.  Further  extension  of  factorable  polynomials. 

EXERCISES 

1.  Multiply  as  indicated  and  make  a  rule  by  which  we 
may  find  by  inspection  the  products  of  polynomials  like  the 
following: 

1.  {x+y){x''-xy+y^)  6.  (3a-6)(9a2+3a64-62) 

2.  {x-y){z''+xy+y'')  7.  (2a+36)(4a2_6a6+962) 

3.  (a+6)(a2-a64-62)  g^  (Sa^ + 562) (9a^_l 50252 +2564) 

4.  (a-6)(a2+a6+62)  9.  (7a3-462)(49a6+28a362+1664) 
6.  (a+26)(a2-2a6+462)  10.  {2a%''-Zc'){4.a'h'JrQa?¥c'^Qc') 

2.  Make  a  rule  for  factoring  the  sum  of  two  cubes. 

3.  Make  a  rule  for  factoring  the  difference  of  two  cubes. 

4.  Factor  64a3-f2763. 

The  expression  is  the  sum  of  two  cubes,  64a3  =  (4a)3  and 
276' =  (36)'.  Therefore,  one  factor  is  the  sum  of  the  cube  roots  of 
64a'  and  276',  i.e.  (4a +36). 

The  other  factor  is  obtained  from  the  first  factor  as  follows: 
square  the  first  term,  (4a)2  =  16a2,  subtract  the  product  of  the  two  terms, 
—  (4a)  (36)  =  —  12a6,  add  the  square  of  the  second  term,  (36)2  =  962, 

Hence,  64a' +276'  =  (4a +36)  (16a2  -  12a6 +962) , 

6.  Factor  8x'- 125?/'. 
Show  by  multiplying  that — 

8x' - 125?/' =  (2a:  -  5?/)  (4a:2  +  lOx?/ +25?/2) . 

Explain  how  the  factor  ^x' +  \Qxy +2by^  may  be  formed  from 
the  terms  of  the  factor  2x  —  5y. 

6.  Form  proportions  from  x^-y^  =  m^-\-n^.     (Apply  §  188.) 

7.  Form  proportions  from  p^—v^  =  a^-b^. 


PROPORTION.    FACTORING.    VARIATION 


87 


Factor  the  following  expressions,  doing  as  many  as  you  can 


mentally: 

8.  a'+b' 

18.  125x^+8y^ 

9.  a'-b' 

19.  27a3+6463 

10.  8x^-2/3 

20.  512c' -27 d' 

11.  m3+27n3 

21.  ii:3p+343 

12.  8c^-d^ 

22.  729a«+216c« 

13.  343+a;3 

23.  {a+by+c' 

14.  ^3+64 

24.  (m4-n)3-a3 

16.  x3+| 

25.  {w+sy-t' 

16.  ax^-Say^ 

26.  (5m-n)3+c3 

17.  216-27a3 

27.  (s+20'+27:i;' 

Proportions  Obtained  from  Given  Proportions 

191.  Proportions  may  be  obtained  from  other  propor- 
tions in  various  ways,  as  is  shown  in  the  following : 


^  EXERCISES 

1.  Using  a  length  equal  to  2  centi- 
meters as  a  unit,  measure  to  two  places  of 
decimals  AB,  DB,  DA,  EB,  EC,  and  BC, 
Fig.  105,  and  show  by  dividing  that — 

BDBE  ^_EC 

^'  DA~EC  ^'  DB~EB 

BD_DA  BA_BC 

^'  BE~EC  BD~BE 

2.  Apply  the  test  of  proportionahty  to  the  following: 


*•  DA' EC 


4     12 

1. 

7"21 

4'     7 

2. 

12~21 

7   ^21 

3. 

4~12 

4+7     12+21 


88  SECOND-YEAR  MATHEMATICS 

3.  What  change  in  the  position  of  the  terms  of  proportion 
1,  exercise  2,  will  transform  it  into  proportion  2?  Equation 
2  is  said  to  be  obtained  from  1  by  alternation. 

4.  What  change  will  transform  proportion  1  into  3  ?  Equa- 
tion 3  is  said  to  be  obtained  from  1  by  inversion. 

6.  What  change  will  transform  proportion  1  into  4  ?  Equa- 
tion 4  is  said  to  be  obtained  from  1  by  addition. 

6.  What  change  will  transform  proportion  1  into  6  ? 

7.  What  change  will  transform  proportion  1  into  6  ?  Equa- 
tion 6  is  said  to  be  obtained  from  1  by  subtraction. 

192.  Alternation.  When,  by  interchanging  the  means, 
or  by  interchanging  the  extremes  of  a  given  proportion 
a  second  proportion  is  formed,  it  is  said  to  be  obtained 
from  the  given  proportion  by  alternation. 


EXERCISES 

1.  Apply  alternation  to  the  proportion  7^=t-6 


a    c 


188 


2.  Apply  alternation  to  r=i 

0    a 

3.  Show  that  if  1=^,  then  -=^  and  f  =  - 

0    a'  c     d         b    a 

Apply  first  the  theorem  of  §  186,  then  of 


4.  Show  that  if    7  =  77  and  -  =  -7,  etc., 
00  c    c 

Fig.  106,  then  -,  =  ^,  =  -,,  etc.  ^ 

^         '  a'    b'    c"  Fig.  lOa 


PROPORTION.    FACTORING.    VARIATION  89 

5.  If  two  equilateral  polygons,  Fig.  107,  have  the  same 
number  of  sides,  the  corresponding  sides  are  in  proportion. 
Prove. 

Show  that   T  =  1  and  w  =  1 

0  0 

•      ^  =  -         Whv? 

■■■  -A   "^'y' 

Similarly,    r,  =  -,=^,,  etc. 

193.  Inversion.  By  inverting  the  ratios  of  a  given 
proportion  a  second  proportion  is  formed,  which  is  said 
to  be  obtained  from  the  given  proportion  by  inversion. 

See  §  191,  exercise  2,  1  and  3. 

EXERCISES 

1.  Apply  inversion  to  r  =  -i . 

2.  Prove  that  if  t  =  j  then  -  =  -  . 

0    a  a    c 

Apply  the  theorems  of  §§  186,  188. 

AB 

194.  Antecedent.     Consequent.     In  a  ratio  as  yrf: , 

or  Ty  AB  and  a  are  the  antecedents  and  CD  and  b  the 
consequents. 

195.  Addition.  Subtraction.  Theorem:  In  a  propor- 
tion the  sum  {or  difference)  of  the  terins  of  one  ratio  is  to 
the  antecedent,  or  consequent,  as  the  sum  {or  difference)  of 
the  terms  of  the  other  ratio  is  to  its  antecedent  or  consequent, 

8     16 
Thus,  from   r  =t7^j  we  obtain  the  proportions 

8+5     16+10       ,  8-5     16-10 

-T-^-ir-^^^^^-io"- 


90  SECOND-YEAR  MATHEMATICS 

The  resulting  proportion  is  said  to  be  obtained  from 
the  given  proportion  by  addition  if  the  sum  is  taken,  and 
by  subtraction  if  the  difference  is  taken. 

^.         a    c 
Given  T=3. 
0    a 

^  ,,    ,    a+&     c-\-d 

To  prove  that  —r-  =—r- . 

Analysis: 

.     .  a-{-b     c+d 

1.  Assume  —r-  =  — r~ 

0  d 

2.  Then  {a+b)d=  {c+d)b        Why? 

3.  .*.     ad+bd  =  cb-\-bd         Why? 

4.  .-.     ad  =  cb  Why? 


6~d 


Why 


The  proof  is  obtained  by  reversing  the  steps  in  the 
preceding  analysis  as  follows : 


1. 

a    c 
b~d 

Why? 

2. 

ad  =  cb 

Why? 

3. 

ad+bd  =  cb+bd 

Why? 

4. 

ia+b)d={c+d)b 

Why? 

F^ 

a-\-b     c+d 

Whv? 

^'  b  d 

Notice  the  method  used  in  obtaining  this  proof.  First,  we 
assume  the  conclusion  to  be  true. 

Then,  by  correct  reasoning,  we  deduce  a  known  fact,  e.g.,  the 
hypothesis.  The  steps  being  reversible,  we  start  from  this  known 
fact  and  get  the  conclusion  by  reversing  the  steps. 

This  last  part  is  the  proof  of  the  theorem. 
Similarly,  prove  that  ~T~  =  ~^  • 


PROPORTION.    FACTORING.    VARIATION  91 

196.  Theorem:  In  a  proportion  the  sum  of  the  terms 
of  one  ratio  is  to  their  difference  as  the  sum  of  the  terms  of 
the  other  ratio  is  to  their  difference. 

Thus,  if  o  =  -7r>  it  follows  that  -r  =  T7i- 
'3     9.  4      12 

EXERCISE 

Use  the  method  of  analysis,  as  in  §  195,  to  prove  that 

.» a     c    ^,       a-\-h     c-\-d 

if  T  =  -i,  then 7  = ;. 

0    d  a—b     c—d 

197.  Addition     and     subtraction.      The     proportion 

r= ;  is  said  to  be  formed  from  t  =  j  by  addition 

a—h    c—d  0     d 

and  subtraction. 

EXERCISES 

Apply  addition  and  subtraction  to  the  following  proportions: 
2m+3n    2s-{-St 


1. 


2m-3n    2s-3t 


2m+Sn+2m-Sn _2s+St-\-2s-3t         4m_4s       _  jn_s 
2m-\-3n-2m+dn~2s+3t-2s+3t'  ^^  Qn  ~Qt '  °^'  n~ t 

^   a    s 

x  —  a-i-b_a—b—x 
x-\-a—b    a+6+a; 

4.     /  /  =o 

V  l+x-v l-x 

Solve  for  x : 

2-\-v'x     Vx-\-5+Vx 


5. 


2-vx     v/x+S-v^a; 

Apply  addition  and  subtraction  and  then  solve. 


92 


SECOND-YEAR  MATHEMATICS 


198.  Theorem:  If  two  or  more  ratios  are  equal,  the 
sum  of  the  antecedents  is  to  the  sum  of  the  consequents  as  any 
antecedent  is  to  its  consequent. 


2    4      8 
Thus,  from  3=^=^2^ 


to  this  theorem,  that 


2+4+8 


3+6+12 


it  follows,  according 

.    .    _2 
.    .    ~3* 


^.         a     c    e    g 

^^^^"^  b^rrh   ■  ■  ■ 

,,    .   a+c+e+g  + 

To  prove  that  ^q:^q:^:p^ 


a_c 
b~d 


Proof 


a_a 

c  _a 
d~h 


^--    etc 

ah  =  ab 
cb  =  ad 
eb  =  af 
gh  =  ah,  etc. 


Why? 
Why? 
Why? 

Why? 

Why? 
Why? 
Why? 
Why? 


Adding  (a+c+e+gf )h  =  a{h+d+f+h ) 


a-\-c-\-e-\-g 

"      h+d+f+h '..      h    d 

EXEKCISES 

Prove  the  following  exercises: 

c2 


a     c      , 
=  jj  etc. 


Why 


/a     Vi 


1.  If  |=J,  it  follows  that  p=^;p-^3;  y^^"/^ 


2a    2c 


ma    mc 


2.  If  T  =  ^,  it  follows  that  3^=3^;  and  ^=^ 


PROPORTION.    FACTORING.    VARIATION  93 

Prove  the  following  by  the  method  of  analysis: 


CL      C 

3.  If  T  =  "7 ,  then 


Sa-\-b    Sc-\-d 


6"^'  b  d 


^^a    c    ,,  a    a+56 

4.  If  ^=5,  then  ~  =  ^q:5^ 


5.  If -  =  7,  then 


S2j+2t    Sx+2s 
y~t'  ^^^"^^^         4y     ~     ix 

^'  ^^  b~d~f''^^^^  b+2d'd+3f~f+5b 
7.  If  ^  =  ^,  then         ^+^=5^ 

Analysis :    Assume      rq— 1  =  r^ 

Then  a¥c +fe2c2  =  a?bd  -^aH^        Why  ? 

Since,  bc  =  ad  Why? 

and  ¥c^  =  a''d^  Why? 

.-.      a¥c  =  o?hd  Why? 

hc  =  ad  Why? 

The  proof  is  obtained  by  retracing  the  steps  in  the  analysis. 


8.  If  7  =-^,  then 


a^^c^    ab^-cd 


i~d'  '"^"  a^-c^~ab-cd 

n    T.«    &    ,,  a'+b'    b'-\-c' 

9.  If  T  =  ",  then      = ■ 

6     c'  a           c 

10.  If  7  =  -,  then      70  ,    g  =  ro 5 

6     c'  ¥-{-c^    b^—c^ 


,,    ^,a    b    ^,  a?+ab    ¥+bc 

11.  If  y  =  -,  then     = 

6     c '  a  c 


if^-.^-^   +1.      c^+c    c+e    e+o^ 


94  SECOND-YEAR  MATHEMATICS 

199.  We  have  seen  in  §§  192-98  that  from  a  propor- 

CL      C 

tion,  as  T  =  -j,  the  following  proportions  may  be  obtained : 


1. 

ah       ^  d    c 

-=-j  and  T  =  - , 
c     d         ha' 

by  alternation 

2. 

h     d 
a    c' 

by  inversion 

3. 

a+h    c-\-d       ,  a-\-h     c-^d 

= and  — T—  =  —1-  , 

a          c               b          d    ' 

by  addition 

4. 

a  —  b     c  —  d       ,  a  —  h     c—d 

= and  —r-=-—^y 

a          c                b          d    ' 

by  subtraction 

5. 

a-\-b     c-{-d 

by  addition  and  sub- 

a-b~c-d' 

traction 

f\ 

a+c    a    c 

by  §  198 :  the  sum  oi 

b+d~bd  the  antecedents  is 

to  the  sum  of  the 
consequents  as  any 
antecedent  is  to  its 
consequent. 

EXEKCISES 

1.  Divide  40  into  parts  that  are  in  the  ratio  of  3 : 5. 

2.  Divide  44  into  parts  in  the  ratio  of  2/3 : 4/5. 

3.  Divide  m  into  parts  in  the  ratio  of  a:c. 

4.  The  denominator  of  a  fraction  is  5  greater  than  the 
numerator,  and  the  value  of  the  fraction  is  2/3.  Find  the 
fraction. 

5.  The  value  of  a  fraction  is  2/3.     If  3  is  added  to  both 

terms  the  value  becomes  7/10.    Find  the  fraction. 

2x 
The  required  fraction  is  of  the  form  ^.    Why  ? 


PROPORTION.     FACTORING.     VARIATION  95 

6.  The  value  of  a  fraction  is  2/5.  If  5  be  added  to  the 
denominator  and  subtracted  from  the  numerator,  the  value 
becomes  3/10.    Find  the  original  fraction. 

t7.  Solve  each  of  the  following  if  the  value  of  the  original 
fraction  is  5/7: 

1 .  If  1  be  added  to  both  terms  the  value  of  the  fraction  becomes 
8/11.     Find  the  original  fraction. 

2.  If  1  be  subtracted  from  both  terms  the  value  becomes  7/10. 
Find  the  original  fraction. 

3.  If  1  be  added  to  the  numerator  and  subtracted  from 
the  denominator  the  value  becomes  4/5.  Find  the  original 
fraction. 

4.  If  1  be  subtracted  from  the  numerator  and  added  to  the 
denominator  the  value  becomes  7/1 1 .    Find  the  original  fraction. 

12    3 

8.  Find  the  values  of  x  and  y  from  -  =    =  ^  . 

^  X     y     1 


Relation  between  Proportion  and  Variation 

200.  Direct  Variation.  When  two  variables  change 
values  but  have  always  the  same  ratio,  each  is  said  to 
vary  directly  as,  or  to  vary  as,  the  other. 

Thus,  the  number  y  is  said  to  vary  directly  as  x,  if 

the  ratio  -  remams  constant,  x  and  y  both  changing,  or 

X 

varying.  .  The  equation 

X 

expresses  algebraically,  and  is  equivalent  to,  the  state- 
ment that  y  varies  directly  as  x 
Show  that  2/  is  a  function  of  x. 


96  SECOND-YEAR  MATHEMATICS 

201.  Relation  between  direct  variation  and  proportion. 

Let  y  vary  directly  as  x  and  let  Xi,  yi;  X2,  2/2;  Xz,  2/3,  etc., 
be  corresponding  values  of  x  and  y. 

Since  y  =  ex,  it  follows  that  -  =  c  and  that  —=^c,—  =  c, 

^  Xl        'X2        ' 

^  =  c,  etc. 

Xz 

Therefore,  ^^=^^ 

Xi      X2 

From  this  equation  we  can  determine  any  one  of  the  four 
numbers  xi,  2/1,  x^  and  2/2,  if  the  other  three  are  given. 

EXERCISES 

1.  The  area  of  a  rectangular  piece  of  land  of  given  width 
varies  directly  as  the  length.  If  the  area  of  a  piece  30  ft.  long 
is  2100  sq.  ft.,  what  must  be  the  length  of  a  strip  containing 
10500  sq.  feet? 

Since  the  area  varies  directly  as  the  length, 
A,  J, 

Tint  ^^     ^' 

^2  =  10500, 
and  Zi  =  30 

2100  _30 
^^^^^y  10500     U  ' 

Solve  this  equation  for  Z2. 

2.  The  cost  of  silk  of  a  certain  grade  varies  as  the  number 
of  yards.     If  35  yd.  of  silk  cost  $61 .  25,  find  the  cost  of  90  yards. 

3.  One  hundred  feet  of  copper  wire  of  a  certain  size  weighs 
35  pounds.    What  is  the  length  of  wire  weighing  175  pounds  ? 

4.  If  y  varies  as  x,  and  if  y  —  SO  when  a;  =  10,  what  is  the 
value  of  y  when  a;  =  18  ? 

202.  Inverse  variation.  When  two  numbers  so  vary 
as  to  leave  the  product  of  ajny  value  of  one  by  the  corres- 
ponding value  of  the  other  constant,  then  one  is  said  to 
vary  inversely  as  the  other. 


PROPORTION.    FACTORING.    VARIATION  97 

The  equation 

xy=c 

expresses  algebraically,  and  is  equivalent  to,  the  state- 
ment that  the  variable  y  varies  inversely  as  the  variable  x. 
Show  that  2/  is  a  function  of  x. 

203.  Relation  between  inverse  variation  and  propor- 
tion. Let  y  vary  inversely  as  x  and  let  Xi,  yi;  X2,  2/2; 
X3, 2/3;  etc.,  be  corresponding  values  of  x  and  y. 

Since  xy  =  c,  it  follows  that  Xiyi  =  c,  X2y2  =  c,  Xzyz  =  c, 
etc. 

Hence,  Xiyi  =  oc2y2' 

From  this  we  may  obtain  the  proportion  —  =^. 
•"  ^    ^  X2    yi 

If  any  three  of  the  four  numbers  xi,  %,  yi,  and  2/2  are 
given,  the  fourth  may  be  found  from  this  proportion. 

EXERCISES 

1.  The  volume  of  air  in  a  bicycle  pump  varies  inversely  as 
the  pressure  on  the  piston.  If  the  volume  is  16  cu.  in.,  when  the 
pressure  is  18  lb.,  what  is  the  pressure  when  the  volume  is 
2  cubic  inches  ? 

2.  The  pressure  of  steam  in  an  engine  cyUnder  varies 
inversely  as  the  volmne.  When  the  pressure  is  100  lb.  per  sq.  in. 
the  volume  is  50  cubic  inches.  What  will  be  the  pressure  per 
sq.  in.  when  the  volume  is  75  cubic  inches  ? 

3.  If  X  varies  inversely  as  y  and  if  a;  =  f  when  y  =  ^,  find  the 
value  of  y  when  x  =  1  J. 

204.  Historical  note.  Like  many  other  mathematical 
topics,  proportion  was  long  used  before  men  comprehended  its 
principles.  The  two  forms  of  proportion  that  have  been  studied 
for  over  two  thousand  years  are  proportion  applied  to  numbers, 
and  proportion  applied  to  line-segments  and  areas. 

Proportion  as  applied  to  numbers  is  one  of  the  oldest  mathe- 
matical topics.     In  the  oldest  known  mathematical  writing,  the 


98  SECOND-YEAR  MATHEMATICS 

Book  of  Ahmes  (see  Cajori,  p.  11),  written  by  an  Egyptian 
scribe  1700  B.C.,  proportion  is  one  of  the  important  subjects. 
The  ancient  Chaldeans,  Phoenicians,  Hindus,  Chinese,  and 
Greeks  all  gave  it  an  important  place  in  their  books.  The 
Greeks,  Arabs,  Hindus,  Moors,  Romans,  and  other  European 
peoples  of  the  dark  and  mediaeval  ages,  that  made  any  preten- 
sions to  learning,  all  emphasized  the  doctrine  of  proportion. 
Mediaeval  geometries  and  mercantile  arithmetics  made  it  a 
major  theme.  Indeed,  until  fifty  years  ago  the  "single  rule 
of  three"  and  the  "double  rule  of  three,"  which  meant  simple 
proportion  and  compound  proportion,  made  up  most  of  advanced 
arithmetic. 

The  principles  of  proportionaUty  as  applied  to  line-segments 
and  to  areas  were  first  studied  by  the  Greeks.  They  drew  their 
beginnings  from  Egypt  and,  perhaps,  Babylon.  Thales  of 
Miletus  (640-546  B.C.)  used  proportionaHty,  perhaps  without 
knowing  it.  The  Pythagoreans  (after  529  B.C.)  employed  it 
more  extensively.  Archytas  of  Tarentum  (428-347  B.C.)  ex- 
tended the  theory  greatly.  Plato  (429—348  b.c.)  was  well 
versed  in  it,  and  Eudoxus  of  Cnidos X408-355  b.c.)  greatly  per- 
fected the  form  of  the  doctrine.  Euclid's  Elements  (300  b.c.) 
devotes  the  fifth  and  a  part  of  the  sixth  book  to  the  doctrine  of 
proportionahty  as  appHed  to  line-segments  and  areas,  a  form  of 
the  doctrine  believed  to  be  due  to  Eudoxus. 

Every  nation  and  people  that  has  acquired  any  standing  in 
mathematics  has  given  great  attention  to  this  doctrine.  It  was 
once  the  most  practical  part  of  all  geometry,  and  some  of  the 
most  practical  subjects  and  topics  of  mathematics  are  still  based 
on  it. 

Summary 

205.  The  chapter  has  taught  the  meaning  of  the 
following  terms : 

alternation  addition  and  subtraction 

inversion  direct  variation 

addition  applied  to  aproportion  inverse  variation 

subtraction  applied  to  a  pro-  antecedent 

portion  consequent 


PROPORTION.     FACTORING.     VARIATION  99 

206.  The  following  theorems  have  been  proved: 

1.  In  a  proportion  the  product  of  the  means  equals  the 
product  of  the  extremes. 

2.  //  the  product  of  two  factors  equals  the  product  of 
two  others,  proportions  may  he  formed  by  taking  as  means 
the  factors  of  one  product  and  as  extremes  the  factors  of  the 
other  product. 

3.  Proportions  may  he  ohtained  from  other  proportions 
by  alternation,  by  inversion,  by  addition,  by  subtraction,  by 
addition  and  subtraction. 

4.  If  two  or  more  ratios  are  equal,  the  sum  of  the  ante- 
cedents is  to  the  sum  of  the  consequents  as  any  antecedent 
is  to  its  consequent. 

207.  The  following  expressions  may  be  factored : 

I.  Polynomials:  having  a  common  factor,  as  ax -{-ay. 

II.  Binomials:  which  are  the  difference  of  two  squares, 
as  x^  —  y'^; 

the  difference  of  two  cubes, 
as  o?  —  W; 

the  sum  of  two   cubes,   as 
a^^b\ 

III.  Trinomials:  which  are  perfect  squares, 

as  x'^=i=2xy-\-y^; 
which  are  of  the  form  ax^-\-bx+c. 

208.  The  relation  between  variation  and  proportion 

has  been  shown. 


CHAPTER  VII 


SIMILAR  POLYGONS 
Uses  of  Similar  Triangles 

209.  Similar  triangles  and  polygons.  We  saw  in  our 
work  of  the  first  year  that  similar  triangles  have  the 
following  two  important  properties: 

(1)  the  ratios  of  the  corresponding  sides  are  equal  and 

(2)  the  corresponding  angles  are  equal. 

The  same  properties  are  possessed  by  similar  poly- 
gons.     For    this 
reason  similar  poly- 
gons are  defined  as 
polygons  having  the 
corresponding  sides 
proportional  and 
the   corresponding 
angles    equal. 
Hence,    the   state- 
ment: polygon  ABCDEFc^A'B'C'D'E'F',  Fig.  108,  may 
be  expressed  symbolically  by  the  two  following  state- 
ments: 


Fig.  108 


a  _  6  _  c  _d  _e  _f 
a'~h'~?~d~e'~f 
2.   ZA  =  ZA',  Z5=ZB',  ZC=ZC',  etc. 


210.  Uses  of  similar  triangles.  Many  problems  may 
be  solved  by  the  aid  of  similar  triangles,  as  may  be  seen 
from  the  following  exercises. 

100 


SIMILAR  POLYGON?? 


im- 


EXERCISES 

1.  To  find  the  height  of  a  chimney. 

Let  AC,  Fig.  109, 
represent  the  shadow 
of  the  chimney  AB, 
and  A'C  the  shadow 
of  a  vertical  stick  A'B'. 

Assuming  rays  of 
sunUght  to  be  parallel, 
show  that   ZC=ZC'. 

Since   triangles 
ABC  and  A'B'C  have  two  angles  equal  respectively ,  they  can  he  shown 
to  he  similar  (§217). 

AB       AC 


Hence, 


and 


A'B'    A'C 
AB=AC 


A'B' 
A'C 


Why? 
Why? 


A  ID 


Using  this  equation  as  a  formula,  find  the  height  of  a  chimney 
whose  shadow  is  108  ft.,  if  at  the  same  time  the  shadow  of  a  4-ft. 
vertical  stick  is  9  ft.  long  ? 

2.  To  determine  the  distance  across  a  river. 

Sighting  across  the  river  with  telescope  A,  Fig.  110,  place  in  the 
line  of  sight 
vertical  rods, 
as  at  .B  and 

C.  Take 
readings  of 
rods  at  E  and 

D.  Depress 
the  telescope 

sighting  at  C  and  take  the  reading  at  F.     From  the  readings  com- 
pute the  length  of  DF  and  EC. 

EC  II  DF        (See  §  373.) 
.'.     Triangles  AFD  and  ACE  are  similar. 

For,  a  line  parallel  to  a  side  of  a  given  triangle  forms  with  the 
other  two  sides  a  triangle  similar  to  the  given  triangle  (§214). 

^  AE    EC  \ 

Hence,     ^7^  =  ^ 

AD 'EC 


Fig.  110 


AD 
AE 


DF 


which 


1.02  SECOND^YEAR  MATHEMATICS 

expresses  AE  in  terms  of  the  known  lengths  AD,  EC,  and  DF. 

The  length,  ED,  may  be 
found  by  subtracting  AD  from  B 

AE. 

3.  To  determine  an  in- 
accessible distance. 

Let  AB,  Fig.  Ill,  be  the 
distance  to  be  measured. 

From  a  point  C,  chosen  con- 
veniently, measure  BC  and  AC. 

Mark  point  D  on  AC. 

On  BC  determine  point  E  so  that  ^rj  =  ttb  •     Measure  DE. 

Triangles  CDE  and  CEB  may  be  shown  to  be  similar. 

For,  two  triangles  are  similar  if  the  ratio  of  two  sides  of  one  equals 

the  ratio  of  two  sides  of  the  other,  and  the  angles  included  between  these 

sides  are  equal  (§218). 

AB_AC 
Hence,     ^^     ^^ 

and  AB  =  — j^ — 

Thus  DE,  AC,  and  DC  being  known,  AB  may  be  found  as  the 
quotient  of  the  product  DE  •  AC  and  DC. 

{211.  To  find  graphically  the  quotient  of  two  arithmetical 
numbers. 

There  are  a  number  of  instruments  for  performing 
mechanically  the  processes  of  multiplication,  division,  and 
extraction  of  roots.  Fig.  112  is  a  device  based  upon 
similar  triangles  for  finding  the  quotients  of  arithmetical 
numbers 

Let  OA  be  the  dividend-line  and  OB  the  line  of  divisors. 

To  divide  42  by  72,  let  the  side  of  a  large  square  repre- 
sent 10. 

Lay  off  00  =  4:2  and  from  C  lay  off  vertically  CD  =  72. 

Stretch  the  string  fastened  at  O  so  that  it  passes 
through  D,  meeting  the  quotient-line,  FQ,  at  E. 


SIMILAR  POLYGONS 


103 


90       100       110     A 


Fig.  112 


104  SECOND-YEAR  MATHEMATICS 


Then  ^ttt:  represents  the  quotient  =^ 
lUU  7  J 

For,  triangles  OFE  and  OLD  are  similar. 

FE    OF 
Therefore,  ^  =  ^ 

Hence,        ^  =  0L        ^^^• 


and 


FE^^^ 
100     72 


42 
Since  FE  =  58  approximately,  it  follows  that  =^  =  .  58, 

approximately. 

EXERCISES 

1.  Using  Fig.    112  find  the  following  quotients  approxi- 

.1    .    .      ^-11  76     64    45    57 

mately  to  two  decimal  places :   t^h,  oo,  q^,  ^q. 

42 

2.  Find  the  quotient  ij^,  using  MP  as  quotient-line. 

Since  "       AOMN  c^  AOLD, 

it  follows  that  y-^  =  YyjT- 

•      ^  =  ^         Why? 
"     OM    OL         ^^^^^ 

MN^42 
10      72 

42  .  1 

Hence,  the  quotient  =^  could  be  obtained  by  taking  —  of  MN 

which  is  .  6  approximately. 

In  a  similar  way  find  the  quotient  ^  . 

3.  In  a  freshman  class  of  130  pupils  taking  mathematics, 
21  obtained  a  grade  of  A,  29  a  grade  of  B,  35  a  grade  of  C,  27  a 
grade  of  D,  and  18  failed.  What  per  cent  of  pupils  in  the  class 
received  a  grade  of  A?  of  B?  of  C?  of  D?  What  per  cent 
failed? 


SIMILAR  POLYGONS  105 


Suppose  X  per  cent  of  the  pupils  receive  an  A  grade. 

21  =  -^ 
100 

X       21 


Then,  21=^^^.130 


H^^^^'  100     130 

21 
From  Fig.  112,  we  find  t^-    16,  approximately. 

Therefore,  approximately  16  per  cent  receive  an  A  grade. 

212.  Construction  of  similar  polygons. 
Let   ABCDEF,    Fig.  ^ 

113,   be   a  given  polygon.  a^^^^^FT ^ — \" 

To  construct  a  polygon       j^'"  |  /   \    "^--^  \     N 
similar  to  A  5  CD  EF. 

Construction:  Draw 
diagonals  from  one  vertex,  as 
B,  to  the  other  vertices,  and       F'^^"--- -^  ^     / 

extend  them.  ^""^-^E' 

From  any  point  on  BA,  as  Pjq  h^ 

A',6iB.wA'F'\\AF. 

Draw  F'E'  II  FE,  E'D'  \\  ED  and  D'C  II  DC. 

Then  A'BC'D'E'F'  is  the  required  polygon. 

Proof:  Prove  ZD=ZD',  AE=^/.E',  etc. 

CTi      TiT)        TiW 
Show  that       'cD'^BD'^^WE"^^^'     ^§  ^^^^ 

CD      DE      EF  „^     ^ 

Hence,  cyD'^WW^^WV"^^^'        ^^y- 

213.  Homologous  parts.     Corresponding  sides  of  simi- 
lar polygons  are  homologous  sides. 

Corresponding  angles,  diagonals,  altitudes,  and  medians 
are  homologous  angles,  diagonals,  altitudes,  and  medians. 

EXERCISES 

1.  Show  that  congruent  polygons  are  similar. 

2.  Show  that  polygons  similar  to  the  same  polygon  are 
similar  to  each  other. 


106  SECOND-YEAR  MATHEMATICS 

Theorems  on  Similar  Figures 

214.  Theorem:  A  line  parallel  to  one  side  of  a  triangle 
forms  with  the  other  two  sides  a  triangle  similar  to  the  given 
triangle. 


Fig.  114 
Given  A  ABC,  smd  DE  \\  AB,  Fig.  114, 
To  prove  that  A  DECc^AABC. 
Analysis:  ^hat  conditions  must  be  satisfied  to  make 

two  triangles  similar?    More  definitely,  what  must  be 

shown  for  triangles  ABC  and  DEC  f 

Proof:    Prove  that  the  angles  of  A  DEC  are  respec- 
tively equal  to  the  angles  of  A  ABC. 

Since  DE  \\  AB         Why  ? 

.      CD    CE         .,„     „ 
••     CA^CB         ^^^' 
Draw  DFWBC. 

Then  g  =  ^.      Why. 

Quadrilateral  DFBE  is  a  parallelogram.      Why  ? 
.-.     DE  =  FB         Why? 
Substituting  for  FB  its  equal,  DE, 
DE^DC 
AB    AC 

This  may  be  written  y^  =x« 

„  CD    CE    DE        ,,,,     „ 

Hence,  CA=CB=AB        ^^y' 

.*.     AABCc^ADEC  Why? 


SIMILAR  POLYGONS  107 

215.  Conditions  sufficient  to  make  triangles  congruent. 

In  geometry  we  have  seen  the  importance  of  congruent 
triangles  in  proving  theorems  and  solving  problems.  The 
definition  of  congruent  triangles  contains  six  conditions,  viz. : 

1.  The  equality  of  the  corresponding  angles, 

ZA  =  ZA',  ZB=ZB\  ZC=ZC\ 

2.  The  equality  of  the  corresponding  sides, 

a  =  a\  b  =  h'j  c  =  c\ 

However,  it  was  shown  that  we  do  not  need  to  estab- 
lish all  of  these  conditions  to  prove  two  triangles  con- 
gruent and  that  the  following  conditions  are  sufficient: 

1.  Two  sides  and  the  angle  included  between  them  in 
one  triangle  equal  respectively  to  the  corresponding  parts 
of  the  other  triangle* 

2.  Two  angles  and  the  side  between  their  vertices 
equal,  respectively. 

3.  Three  sides  equal,  respectively. 

Thus,  the  problem  of  proving  two  triangles  congruent 
is  greatly  simplified. 

216.  Conditions  sufficient  to  make  two  triangles 
similar. 

The  definition  of  similar  triangles  contains  five  con- 
ditions, viz. : 

1.  The  equality  of  the  corresponding  angles,  or — 

ZA=ZA',  ZB=AB\  ZC=ZC\ 

2.  The  proportionaUty  of  the  corresponding  sides,  or 

-,=T/i  T,  =  -fy  from  which  it  follows  that  -,=-,. 
a'    6"  b     c  c     o! 

As  in  the  case  of  congruent  triangles  it  is  not  necessary 

to  show  that  all  five  of  these  conditions  are  satisfied  to 


108  SECOND-YEAR  MATHEMATICS 

make  two  triangles  similar.  It  will  be  shown  that  any 
one  of  the  following  three  conditions  is  necessary  and 
sufficient: 

1.  The  equality  of  two  pairs  of  corresponding  angles. 

2.  The  proportionality  of  two  pairs  of  corresponding 
sides,  and  the  equality  of  the  included  angle. 

3.  The  proportionality  of  the  corresponding  sides. 

217.  Theorem:     Two    triangles    are    similar    if   two 
angles  of  one  are  respectively  equal  to  two  angles  of  the  other. 


Given  A  ABC  and  A'B'C,  with  A=A'  and   C=C. 
Fig.  115. 

To  prove  that  A  ABCc^A  A'B'C. 


Proof: 

STATEMENTS 

REASONS 

On  C'A' 

layoff  CD  =  CA 

Draw 

DE  11  A'B' 

Then, 

ADEC'c^AA'B'C 

§214 

ADEC'^AABC 

a.s.a. 

.-.     AABCc^AA'B'C 

Why? 

EXERCISE 


Two  right  triangles  are  similar  if  an  acute  angle  of  one  is 
equal  to  an  acute  angle  of  the  other. 


SIMILAR  POLYGONS 


109 


218.  Theorem:  Two  triangles  are  similar j  if  the  ratio 
of  two  sides  of  one  equals  the  ratio  of  two  sides  of  the  other 
and  the  angles  included  between  these  sides  are  equal. 


A'                                         B'       A 

B 

Pig.  116 

Given     M.ABC     and     A'B'C, 

with     C^C 

CA      CB     ^.     ,,^ 

C'A'  =  C'B'^  ^'^'  ^^^• 

To  prove  that   AABCc^  AA'B'C. 

Proof: 

STATEMENTS 

REASONS 

OnC'A'layoff  C'D  =  CA 

By  construction 

OnC'B'layoffC'E  =  CB 

By  construction 

inen,                 C'A'~C'B' 

Why? 

.-.     DEWA'B' 

Why? 

.-.     ADEC'c^AA'B'C 

Why? 

But              ADEC'^AABC 

s.a.s. 

.'.    AABCc^AA'B'C 

Why? 

and 


EXERCISES 

1.  Two  right  triangles  are  similar  if  the  ratio  of  the  sides 
including  the  right  angle  of  one,  is  equal  to  the  ratio  of  the  cor- 
responding sides  of  the  other. 

2.  Lines  drawn  joining  the  midpoints  of  the  sides  of  a 
triangle  form  a  triangle  which  is  similar  to  the  first  triangle. 

3.  Two  isosceles  triangles  are  similar,  if  an  angle  in  one  is 
equal  to  the  corresponding  angle  in  the  other. 


110 


SECOND-YEAR  MATHEMATICS 


219.  Theorem:    Two  triangles  are  similar  if  the  cor- 
responding sides  are  in  proportion. 


FiQ.  117 


AB      BC      CA 


driven  /a 

,  ji.n  u  ana  ii  i5  u  ,  will 

A'B'    B'C    CA" 

Fig.  117, 

To  prove 

'.  that  A  ABC  c^  A  A'B'C\ 

Proof: 

STATEMENTS 

REASONS 

On  CA' 

lay  off  C'i)  =  C^ 

On  C'B' 

lay  off  C'^  =  C5 

Then 

CD      C'E 
C'A'~C'B' 

Why? 

.-.     ADECc^AA'B^' 

(§  218) 

CD      DE 
"     CA'    A'B' 

Why? 

But 

CD      AB 
CA'    A'B' 

Why? 

DE      AB 
"     A'B'    A'B' 

Why? 

DE  =  AB 

Why? 

.\     ADEC^AABC 

s.'s.s. 

,\      AABCc^AA'B'C 

Why? 

PROBLEMS   AND   EXERCISES 

Prove  the  following  exercises: 

1.  Two  triangles  are  similar  if  the  corresponding  sides  are 
parallel,  or  perpendicular. 


SIMILAR  POLYGONS 


111 


For,  if  the  sides  of  the  angles  are  parallel  or  perpendicular, 
each  to  each,  the  angles  are  either  equal  or  supplementary. 

Thus,  (1)  A=A',  or(2)  A +A'= 2  right  angles 
(3)  B=B',  or  (4)  ,5 +5' =2  right  angles 
(5)   C  =  C",  or  (6)   C+C'  =  2  right  angles 

Show  that  the  three  equations  (2),  (4),  and  (6)  cannot  all  be 
true  at  the  same  time. 

Show  that  two  of  the  equations  (2),  (4),  and  (6)  cannot  both  be 
true  at  the  same  time. 

Hence,  at  least  two  of  the  equations  (1),  (3),  and  (5)  must  be 
true  and  the  triangles  are  mutually  equiangular. 

Apply  §217. 

|2.  Two  parallelograms  are  similar  if  an  angle  in  one  is 
equal  to  an  angle  in  the  other  and  the  including  sides  are 
proportional. 

J3.  Two  rectangles  are  similar  if  the  ratio  of  two  consecutive 
sides  of  one  is  equal  to  the  ratio  of  the  corresponding  sides  of  the 
other. 

4.  The  perimeters  of  similar 
triangles  are  to  each  other  as  any 
two  homologous  sides. 

Since  the  triangles.  Fig.  118, 
are  similar, 

a  _h  _c 
a'~h'~c' 


Fig.  118 


a'+h'+c' 


5.  The  perimeters  of  similar 
polygons  are  to  each  other  as  any 
two  homologous  sides. 

Since  the  polygons,  Fig.  119,  are  similar 


g+b+c+etc. 
a'+h'+c'+etc- 


=  —, ,  etc. 


r}  =  -,,  etc. 
6'    c  ' 


Why? 
Why? 


112 


SECOND-YEAR  MATHEMATICS 


6.  The  homologous  altitudes  of  similar  triangles  are  to  each 
other  as  the  homologous  sides,  and  as  the  perimeters. 
Prove 

AADCco  AA'D'C,  Fig.  120. 
b      h 


Then 
But 


D  B 


Fig.  121 


5^  feet. 


J7.  The  altitudes  of  a  triangle  are  inversely  proportional 
to  the  sides  to  which  they  are  drawn. 
Prove  ADBCc^  AABE,  Fig.  121. 

J8.  The  homologous  medians  of 
two  similar  triangles  are  to  each  other 
as  any  two  homologous  sides,  and  as 
the  perimeters. 

J9.  The  bisectors  of  homologous 
angles  of  similar  triangles  are  to  each  other  as  two  homologous 
sides,  and  as  the  perimeters. 

JlO.  The  length  of  the  shadow  cast  by  a  4-ft.  vertical  rod  is 
At  the  same  time  the  length  of  the  shadow  cast  by  a 
spire  is  220  feet.     How  high  is  the  spire  ? 

Jll.  A  man  at  a  window  sees  a  point  on  the  ground  in  line 
with  the  top  of  a  post  and  window-sill.  He  finds  that  the  point 
is  2  ft.  8  in.  from  the  foot  of  the  post,  and  that  the  post  is  3  ft. 
high  and  24j  ft.  from  a  point  just  under  the  window.  How  high 
is  the  window  from  the  ground  ? 

tl2.  A  boy  wishes  to  know 
how  far  it  is  from  the  shore  of  a 
lake  at  A  to  an  island,  B,  Fig.  122. 
At  C,  20  yd.  from  A  on  the  line 
BA,  he  lays  off  CD±CB  and 
CD  =  60  rods.  At  A  he  constructs 
a  line  perpendicular  to  AB  meet-  Fig.  122 


SIMILAR  POLYGONS 


U3 


ing  DB  at  E.    By  measuring  he  finds  AE=hQ  rods.     Find  the 
required  distance. 

J13.  The  Hne  joining  the  midpoint  of  one  of  the  bases  of  a 
trapezoid  to  the  point  of  intersection  of  the  diagonals  bisects 
the  other  base. 

14.  The  lengths  of  the  sides  of  a  triangular  piece  of  land  are 
approximately  125  rd.,  54  rd.,  and  112  rods.  A  drawing  is 
made  of  it,  the  longest  side  of  which  is  3  feet.  What  are  the 
lengths  of  the  other  sides  of  the  triangle  in  the  drawing  ? 

15.  The  non-parallel  sides  of  a  trapezoid  of  bases  18  and  60 
and  of  altitude  6  are  produced  until  they  meet.  What  are  the 
altitudes  of  the  triangles  on  the  bases  of  the  trapezoid  ? 

|16.  The  base  of  a  triangle  is  72  in.,  and  the  altitude  is 
12  inches.  Find  the  upper  base  of  the  trapezoid  cut  off  by  a  line 
parallel  to  the  base  and  8  in.  from  it. 

Xn.  Two  sides  of  a  triangle  are  14  in.  and  3.5  in.  and  the 
included  angle  is  75°.  Two  sides  of  another  triangle  are  20  in. 
and  5  in.  and  the  included  angle  is  75°.  Show  that  the  triangles 
are  similar. 

18.  The  perimeter  of  a  triangle  is  15  cm.,  and  the  sides  of  a 
similar  triangle  are  4 . 5  cm.,  6 . 4  cm.,  and  7 . 1  centimeters.  Find 
the  lengths  of  the  sides  of  the  first  triangle. 

19.  The  perimeters  of  two  similar  triangles  are  x^-\-Zx-\-2 
and  16,  and  a  pair  of  homologous  sides  are  respectively  Zx  and  8. 
Find  the  value  of  x. 

$20.  The  perimeters,  p  and  y',  of  two  similar  triangles,  and 


V 

p' 

a           \           a' 

a:2+l 

X 

2i        1        1 

ZK^-\-QK 

27 

35 

4 

4 

y'-2y+l 

1 

4 

a  pair  of  homologous  sides,  a  and  a',  are  expressed  in  the  table 
above.     Find  the  values  of  x,  y,  and  K. 


114 


SECOND-YEAR  MATHEMATICS 


220.  Theorem:  Similar  polygons  may  be  divided  by 
homologous  diagonals  into  triangles  similar  to  each  other 
and  similarly  placed. 


A'       of     B 


Fig.  123 


Given  polygon  ABCD,  etc.,  c^  polygon  A'B'C'D',  etc., 
Fig.  123,  with  diagonals  drawn  from  A  and  A'. 
To  prove  Al^AV,  All ^  All',  etc. 

Proof: 


STATEMENTS 

REASONS 

a     b 
a~b' 

Why? 

B  =  B' 

Why? 

Alc^AV 

Why? 

x  =  x' 

Why? 

C^C 

Why? 

y=y' 

Why? 

b    d 
b~d' 

Why? 

b     c 
b~c' 

Why? 

c      d 
c'    d' 

Why? 

AlIc/5  All',  etc. 

Why? 

SIMILAR  POLYGONS  115 

SummaLTy 

221.  The  following  theorems  were  proved  in  this 
chapter : 

1.  A  line  parallel  to  one  side  of  a  triangle  forms  with 
the  other  two  sides  a  triangle  similar  to  the  given  triangle. 

2.  Two  triangles  are  similar  if  two  angles  of  one  are 
respectively  equal  to  two  angles  of  the  other. 

3.  Two  triangles  are  similar,  if  the  ratio  of  two  sides  of 
one  equals  the  ratio  of  two  sides  of  the  other  and  the  angles 
included  between  these  sides  are  equal. 

4.  Two  triangles  are  similar  if  the  corresponding  sides 
are  in  proportion. 

5.  The  perimeters  of  similar  polygons  are  to  each  other 
as  any  two  homologous  sides. 

6.  Similar  polygons  may  he  divided  by  homologous 
diagonals  into  triangles  similar  to  each  other  and  similarly 
placed. 

222.  It  was  shown  how  to  construct  a  polygon  similar 
to  a  given  polygon. 

223.  Quotients  of  arithmetical  numbers  may  be 
found  mechanically  by  means  of  squared  paper  and  a 
string. 


CHAPTER  VIII 


RELATIONS     BETWEEN     THE     SIDES     OF     TRIANGLES. 
THEOREM  OF  PYTHAGORAS  AND  ITS  GENERAL- 
IZATIONS.    QUADRATIC  EQUATIONS. 
RADICALS. 

Similarity  in  the  Right  Triangle 

224.  Theorem:  The  perpendicular  to  the  hypotenuse 
from  the  vertex  of  the  right  angle  divides  a  right  triangle  into 
parts  similar  to  each  other  and  to 
the  given  triangle. 


Given  AABC  with  the 
right  angle  C,  and  CD±  AB, 
Fig.  124. 

To  prove 


Fig.  124 


AADC^  ABDC^  AABC. 


Proof: 


x  =  x 


Why? 
y  =  y'  Why? 

.-.     AADC^ABDC        Why? 
Prove  that   ^ADC  and  ABC  are  mutually  equi- 
angular and  therefore  similar.  ^ 
Similarly,  prove  ABDC^  AABC. 


225.  Projection  of  a  point.    The  pro-      ^ ^        ^ 

jection  of  a  point  upon  a  given  line  is  the  -piQ.  125 

foot   of   the  perpendicular  drawn  from 
the  point  to  the  hne.     Thus,  point  D,  Fig.  125,  is  the  pro- 
jection of  point  A  upon  BC. 

116 


TRIANGLES.    QUADRATICS.    RADICALS 


117 


226.  Projection  of  a  segment.  To  project  a  line- 
segment,  as  AB,  Fig.  126,  upon  a  line,  as  CD,  drop 
perpendiculars  to  CD  from  the 
endpoints  of  the  segment  AB. 
Then  EF  is  the  projection  of  AB 
upon  CD.  cF 

In  general,  the  projection  of 
a  given  segment  upon  a  line  is 

the  segment  of  the  line  whose  endpoints  are  the  pro- 
jections of  the  endpoints  of  the  given  segment. 


E  F 

Fig.  126 


EXERCISES 


1.  In  each  of  the  following  figures  name  the  projection  of 
AB  upon  CD,  (Figs.  127-29.) 


C      E  F      D 

Fig.  127 


Fig.  129 


Draw  a  figure  in  which  the  segment 
is  equal  to  the  projection. 

2.  In  triangle  ABC,  Fig.  130,  name 
the  projection  of  AC  upon  AB;  of  BC 
upon  AB. 

3.  In  triangle  ABC,  Fig.  130,  project 
BC  upon  AC;  AB  upon  BC. 

4.  Draw  an  obtuse  triangle,  as  ABC, 
Fig.  131.  Project  AB  upon  BC;  AC 
upon  AB;  BC  upon  AB;  AB  upon 
AC. 


Fig.  131 


118  SECOND-YEAR  MATHEMATICS 

227.  Mean   proportional.     In   the  proportion  ?=-, 

0      c 
6  is  a  mean  proportional  between  a  and  c. 


EXERCISES 

1.  Find  a  mean  proportional  between  4  and  9. 


4    X 
Denoting  the  mean  proportional  by  x,  we  have  -=- 

.-.  a;2=4.9  Why? 

.-.  a;  =  ±1/479 

.-.  a;  =±2 -3  Why? 

or  a;=+6,  —  6.  Check  both  results. 

2.  In  triangle  A5C,  Fig.  132,  find    ^ 
the  projection  of  the  median,  m,  upon 
AB. 

228.  Radical.  An  indicated  root  of  a  number  is  a 
radical.     Thus,  1/5,  vx,  F  16,  Va-{-¥  are  radicals. 

229.  Simplification  of  radicals.  In  computing  the 
value  of  a  radical  it  is  often  of  advantage  to  change  the 
form  of  the  number  under  the  radical  sign.  The  following 
examples  illustrate  this: 

J   / 1/25^  =  5  •  4,  for  (5  •  4)(5  •  4)  =  25  •  16. 
•  I1/36  •  9  =  6  •  3,  for  (6  •  3)(6  •  3)  =36  •  9. 

Thus  the  values  of  1^25  •  16,  t/36  •  9,  etc.,  are  found 
by  extracting  the  square  roots  of  the  factors  separately 
and  then  multiplying  the  results. 

In  general,  the  square  root  of  a  product,  as  ah,  may  he 
found  hy  taking  the  square  roots  of  the  factors,  as  a  and  h,  and 
then  taking  the  product  of  these  square  roots.  This  may  be 
stated  briefly  in  the  form  of  an  equation,  thus. 


TRIANGLES.    QUADRATICS.    RADICALS  119 

This  principle  enables  us  to  obtain  by  inspection  the 
square  roots  of  some  large  numbers,  as  is  shown  by  the 
following  examples : 

ri/3136  =  T/4  •  784  =  1/4  •  4  •  196  =  1^4  •  4  •  4  •  49 
11.  \     =2  ♦  2-  2-7  =  56. 

[1/4225  =  1/5  •  845  =  1/5  •  5  •  169  =  5  •  13  =  65 

The  principle  explained  above  may  be  appUed  to 
advantage  even  when  the  number  under  the  radical  sign 
is  not  a  square.     For  example: 


III.  1/50  =  1/5-  5  .  2  =  51/2. 

KJiowing  the  square  root  of  2  to  be  1 .  414+ .  ■ 

it  follows  that  V 50  =  7. 070+   

Similarly,  i/g^  =  y  4a^ .  2a  =  2aV2a 

and     1/108  =  1/902  =  1/9  •  4  •  3  =  6l/3 


EXERCISES 

1.  Reduce  the  following  radicals  to  the  simplest  form: 

1.  1/75  5-  ^1280252  9.  Vo?+2ah^¥ 

2.  ^27  6.  y'lQ2xY  10.  i^4a2-20a6+2562 

3.  v/^  7.  /243a62  n.  i/g^^Tg^ 

4.  V20xhj  8.  #"16  12.  V  (a-\-b)(a''-b^) 

2.  Find  the  mean  proportionals  between  2  and  18;    10  and 
90;  8  and  200;  20  and  180. 

3.  Find  the  mean  proportionals  between  a^  and  b^;  c^  and  d-. 

4.  Find  the  mean  proportional  between  x'^-\-2xy-{-y^  and 
x^—2xy-\-y^. 

6.  Show  that  the  mean  proportional  between  a  and  h  is 
the  square  root  of  the  product  of  a  and  6. 


120 


SECOND-YEAR  MATHEMATICS 


m  D 


230.  Theorem:  In  a  right  triangle,  the  perpendicular 
from  the  vertex  of  the  right  angle  to  the  hypotenuse  is  the 
mean  proportional  between  the  segments  of  the  hypotenuse. 

That  is,  we  are  to  prove  "t  =- »  Fig.  133. 

To  prove  this  proportion,  use 
the  principle  that  in  similar  triangles 
the  sides  opposite  equal  angles  are 
homologous  sides  and  are  therefore 
proportional.  Fig.  133 

231.  Section  230  affords  a  way  of  finding  geometrically 
the  mean  proportional  between  two  segments  (see  prob- 
lem 1,  below). 

Problems  of  Construction 

1.  To  construct  a  mean  proportional  between  two  segments. 

Given  the  segments  m  and  n,  Fig.  134, 
Required    to  ^  ^ 

construct  the  mean  ^^        ~4/^ 

proportional  be- 
tween m  and  n. 

Construction: 

On  a  line,  as  AB,  lay 
.off  AC  =  m,  CD  =  n. 

Draw  CE±AB. 

Draw  the  circle 
AFD  on  AD  as  a  diameter,  meeting  CE  at  F. 

Then  FC  is  the  mean  proportional  between  m  and  n. 

Proof:    Draw  AF,  DF,  and  the  median  HF. 
Show  that  ZAFH=ZA=x 
Show  that  ZHFD=  ZD  =  y 
Then  2x-\-2y  =  180°         Why  ? 

.-.     Z  AFD  =  90°  Why? 

rri__FC 
'''    FC~  n 


Fig.  134 


Why? 


TRIANGLES.    QUADRATICS.    RADICALS 


121 


2.  Construct  a  square  equxil  to  a  given  rectangle. 

Let  a  and  h  be  the  dimensions  of  the  given 
rectangle,  Fig.  135. 

Construct  the  mean  proportional  between  a 
and  h. 


Fig.  135 


On  the  mean  proportional  between  a  and  h 
as  a  side,  construct  a  square. 

Prove  that  the  area  of  this  square  is  equal  to  the  area  of  the  given 
rectangle. 

3.  Construct  the  square  root  of  a  number. 

1.  To  find  the  square  root  of  2,  lay  off  on  squared  paper  two 
factors  of  2,  as  2  and  1,  Fig.  136,  in  the  same  way  as  m  and  n, 
problem  1.     (Use  the  scale  1=2  cm.) 


n± :r..^± 

^'^                                           s 

7                                                   s_ 

_    _       ,Z               -    -                    -            ^^ 

S 

/              .Ti-                         s 

'^                                                                                                                                                              L 

t                ~                                \ 

^ _       _  jr 

.               _ ^ 

/                                                          A 

t 

it 

"IX jT "b ¥                   ^ 

_±    _     ".        ±                   ±      -             -L^ 

Fig.  136 

The  mean  proportional,  BD,  between  AB  and  BC,  represents 
graphically  the  required  square  root  of  2.     Why  ? 
Measure  BD  to  two  decimal  places. 
Check  by  extracting  the  square  root  of  2  to  two  decimal  places. 

2.  Find  geometrically  the  square  root  of  6;  of  5;  of  8. 


EXERCISE 


A  perpendicular  to  a  diameter  of  a  circle  at  any  point,  extended 
to  the  circle,  is  the  mean  proportional  between  the  segments  of  the 
diameter.    Prove. 


122  SECOND-YEAR  MATHEMATICS 

Relations  of  the  Sides  of  a  Right  Triangle 

232.  Theorem :   In  a  right  triangle  either  side  of  the  right 

angle  is  a  mean  proportional  between  its  projection  upon  the 

hypotenuse  and  the  entire  hypotenuse. 

„,  ^  m    a       ,  n    b 

We  are  to  prove    —  =-  and  t  =-  , 
a     c  be 

Fig.  137. 

To  prove  this,  apply  the  principle    '^ '■ — c- 

that  homologous  sides  of  similar  Fig.  137 

triangles  are  in  proportion. 

This  theorem  enables  us  to  obtain  a  proof  for  one  of 
the  most  important  theorems  of  geometry: 

233.  Theorem  of  Pythagoras.  The  square  of  the 
hypotenuse  in  a  right  triangle  is  equal  to  the  sum  of  the 
squares  of  the  sides  of  the  right  angle. 


and, 

and 

.-.     a^-{-¥={m-\-n)c  Why? 

or  a'-hb'=c^ 

The  last  four  steps  in  this  proof  suggest  the  following 
geometric  illustrations : 

The  equation,  a'^  =  m  '  c,  means  that  the  square  on  BCj 
Fig.  138,  is  equal  to  a  rectangle  of  dimensions  m  and  c, 
as  BEFD.  (Notice  that  the  sides  of  the  rectangle 
BEFD  are  m,  the  projection  of  BC  on  AB,  and  BE  which 
is  equal  to  c,  the  length  of  the  hypotenuse  AB.) 

Similarly,  b'^  =  n  -  c  means  that  the  square  on  AC  is 
equal  to  a  rectangle  as  FHAD,  having  the  dimensions 


^      ^    m     a 

Proof:  — =- 

a      c 

Why? 

n    b 
b~c 

Why? 

.'.     a^  =  mc 
¥  =  nc 

Why? 

Why? 

TRIANGLES.    QUADRATICS.    RADICALS 


123 


equal  to  n,  the  projection 
of  BC  on  AB,  and  BH 
which  is  equal  to  the 
hypotenuse,  c. 

Hence,  the  sum  of 
the  squares  on  AC  and 
BC  is  equal  to  the  sum' 
of  these  two  rectangles, 
or  to  the  square  on  the 
hypotenuse. 

This  illustration  may 
be  used  as  an  outline  of 
Euclid's  proof  of  the 
theorem  of  Pythagoras 
given  in  §  462. 

234.  Historical  note :  It  is  said  that  Pythagoras,  jubilant 
over  his  great  accomplishment  of  having  found  a  proof  of  the 
theorem,  sacrificed  a  hecatomb  to  the  muses  who  inspired  him. 
The  invention  was  well  worthy  of  this  sacrifice,  for  it  marks 
historically  the  first  conception  of  irrational  numbers.  It  is 
believed  that  Pythagoras  showed  the  existence  of  irrational 
numbers,  by  showing  that  the  hypotenuse  of  a 
certain  isosceles  right  triangle  is  equal  to  i/2 
(See  Figure.) 

His  followers  found  much  pleasure  in  finding 
special  sets  of  integral  values  of  a,  6,  c  satisfying 
the  equation  a^-^-lf^c^,  the  simplest  set  being 
3,  4,  and  5.     Such  numbers  are  called  Pytha- 
gorean nhmhers.    The  question  naturally  arose  later  whether 
there  existed  any  sets  of  integral  values  of  a,  b,  and  c  that  would 
satisfy  the   equations  a^4-6^  =  c^,  a'^-{-¥  =  c*,   etc.,   in    general, 
a^-\-b^  =  c^  forn>2. 

The  great  mathematician  Fermat,  who  Hved  1601-65,  states 
among  his  notes  the  theorem  that  the  equation  x^-\-y^  =  z^  is 
not  satisfied  by  a  set  of  integral  numbers  for  x,  y,  z,  and  n  except 


124 


SECOND-YEAR  MATHEMATICS 


for  n  =  2.  He  also  makes  the  statement  that  he  has  discovered 
a  really  wonderful  proof  for  the  theorem.  Unfortunately,  he 
gives  not  the  least  suggestion  as  to  the  nature  of  his  proof. 
The  theorem  is  very  simple,  but  it  has  been  impossible  to  this 
day  to  find  a  proof,  although  a  price  of  100,000  marks  ($20,000) 
has  been  offered  by  a  German  society  to  the  fortunate  person 
who  first  gives  a  complete  proof  of  the  theorem,  or  who  shows 
by  a  single  exception  that  the  theorem  is  not  true.  (See  Ball's 
Mathematical  Recreations,  4th  ed.,  1905,  pp.  37-40.) 

EXERCISES 

1.  In  triangle  ABC,  Fig.  139,  ZACB  is  a  right  angle  and 
CD±AB.    AD=2,  Z)5  =  30.    Find  the  lengths  of  AC  and  CB. 

2.  The  radius  of  a  circle  is  12.5,  Fig. 


140.  Find  the 
AC  upon  the 
diameter  AB 
passing  through 
one  of  the  end- 
points  of  the 
chord. 

3.  In    the 

right  triangle 
ABC,  Fig.  141, 
a=12  and  6  =  5. 


projection   of 


find 


Find  b,  m,  n,  and  h,  if  a  =  S  and 


Fig.  141 


m  =  9|-  and 


c  =  10. 

Find  a,  b,  c,  and  h  if 
«  =  5f. 

4.  Compute  the  dimensions  of  the 
section  of  the  strongest  beam  that  can 
be  cut  from  a  cylindrical  log. 

Let  the  circle,  Fig.  142,  represent  a 
cross-section  of  the  log.  Then  the  dimen- 
sions of  the  strongest  beam  are  computed 
as  follows: 

Trisect  the  diameter  AB  at  C  and  D  (§176,  exercise  7). 


PIERRE      DE      FERMAT 


PIERRE  DE  FERMAT  was  born  near  Toulouse  in 
1601  and  died  at  Castres  in  1665.  The  great  mathe- 
matical historian  Cantor  and  others  have  called 
Fermat  "the  greatest  French  mathematician  of  the  seven- 
teenth century,"  and  this  was  a  century  of  great  French 
mathematicians.  He  was  the  son  of  a  leather  merchant 
and  was  educated  at  home.  He  studied  law  at  Toulouse 
and  in  1631  became  a  councilor  of  the  Parliament  of 
Toulouse.  He  is  said  to  have  performed  the  duties  of  his 
office  with  scrupulous  accuracy  and  fideUty.  He  loved 
mathematical  study^  made  it  his  avocation,  spending  most 
of  his  leisure  on  it.  His  disposition  was  modest  and  retiring. 
He  published  very  little — only  one  paper — during  his  life- 
time. Though  his  vocation  was  that  of  a  lawyer  and  parlia- 
mentarian, his  celebrity  rests  upon  what  he  accompUshed 
in  his  avocation. 

Notwithstanding  the  fact  that  Fermat  published  very 
little,  he  exerted  a  great  influence  on  the  mathematicians 
of  his  age  through  a  continual  correspondence  which  he 
carried  on  with  them.  The  mathematical  discoveries  upon 
which  his  fame  rests  were  made  known  to  the  world 
through  his  correspondence  or  through  the  notes  on  his  re- 
sults that  were  found  after  his  death, written  on  loose  sheets 
of  paper,  or  scribbled  on  the  margins  of  books  he  had 
annotated  while  reading.  A  part  of  these  notes  and  Fer- 
mat's  marginal  notes,  found  in  his  copy  of  Diophantus' 
Arithmetic,  were  published  after  his  death  by  his  son, 
Samuel.  As  Fermat's  notes  do  not  seem  ever  to  have 
been  intended  for  pubHcation,  it  is  often  difficult  to  estimate 
when  his  discoveries  were  made,  or  whether  they  were 
really  original. 

Most  of  his  proofs  are  lost,  and  probably  some  of 
them  were  not  rigorous.  He  seems  to  have  worked  care- 
lessly, or  at  least  unsystematically,  for  one  of  his  marginal 
notes  on  an  important  theorem  that  still  awaits  proof, 
notwithstanding  the  facts  that  several  of  the  world's 
greatest  mathematicians  have  tried  their  wits  upon  it  and 
that  the  Paris  Academy  of  Sciences  has  on  two  occasions 
at  least,  in  1850  and  1853,  offered  to  the  world  a  prize  of 
3,000  francs  for  a  complete  proof  of  it,  is  the  remark:  ''I 
have  found  for  this  a  truly  wonderful  proof,  but  the  margin 
is  too  small  to  hold  it." 

The  theorem  referred  to  in  the  foregoing  remarks  is 
called  the  ''greater  Fermat  theorem,"  or  the  "last  Fermat 
theorem"  (see  §  234  of  this  book).  Several  of  Fermat's 
theorems  have  been  proved  by  later  mathematicians,  but 
they  have  required  good  mathernatical  abiUty.  Some  are 
still  awaiting  mathematical  genius. 

For  fuller  information  about  Fermat  see  Ball's  History, 
pp.  293-301;  Cajori's  History,  pp.  173  and  179-82;  or 
Historical  Introduction  to  Mathematical  Literature,  by 
G.  A.  Miller,  published  by  Macmillan. 


TRIANGLES.    QUADRATICS.    RADICALS 


125 


Erect  CE±AB  and  DF±AB. 
Draw  the  quadrilateral  AFBE. 
This  is  the  required  section  of  the  strongest 
beam. 


If  the  diameter  of  the  log  is  15  in. 
AE  and  AF, 


compute 


5.  Prove  that  the  diagonal  of  a  square  is 
equal  to  the  product  of  the  side  by  the  square 
root  of  2,  Fig.  143. 

6.  Prove  that  the  diagonal  of  a  rectangle  is 
equal  to  the  square  root  of  the  sum  of  the  squares 
of  two  consecutive  sides,  Fig.  144. 

7.  Express  the  altitude  of  an  equilateral 
triangle  in  terms  of  the  side,  Fig.  145. 

8.  Fig.  146  represents  a  circular  window. 
The  radius  of  the  largest  circle  is  6.  Find  the 
radius,  x,  of  the  smallest  window. 

The  sides  of  the  right  triangle  ABC 
are  3,  a: +3  and  Q—x  respectively.    Why  ? 

.'.     (a:+3)2  =  (6-a:)2+9 
x2 +6x +9  =  36  -  12a; +a;2 +9 
.-.    a:  =  2 


FiQ.  143 


Fig.  144 


Fig. 145 


A.    3     B 

Fig.  146 


Quadratic  Equations* 

235.  Summary  of  methods  of  solving  quadratic 
equations.  In  the  preceding  course  quadratic  equations 
have  been  solved  by  the  following  three  methods: 

(1)  By  graph. 

(2)  By  factoring. 

(3)  By  completing  the  square. 


*  See  historical  note,  §  238. 


126  SECOND-YEAR  MATHEMATICS 

The  graphical  method  exhibits  to  the  eye  the  solutions 
of  the  equation  and  enables  one  to  determine  the  solutions 
approximately. 

The  method  of  factoring  is  brief,  but  fails  when  we 
are  unable  to  factor  the  trinomial. 

The  method  by  completing  the  square  always  gives 
the  exact  results.  The  objection  to  it  is  the  length  of 
the  process. 

For  this  reason  another  method  will  be  developed 
which  is  not  only  brief,  but  which  can  be  applied  to  any 
quadratic  equation. 

All  quadratic  equations  in  one  unknown  may  be  ar- 
ranged in  the  normal  form  • 

ax''+bx+c=0, 

where  a  stands  for  the  coefficient  of  x^  when  all  terms  in  x^ 
have  been  combined  into  one;  b  denotes  the  coefficient  of 
X,  and  c  is  the  constant,  i.e.,  the  term  or  the  sum  of  terms 
not  containing  x. 

Thus,  in  5x2-f  3a;-4  =  0,  a  =  5,  6  =  3,  c=-4. 

EXERCISES 

Arrange  each  of  the  following  equations  in  the  normal  form, 
ax^+bx-\-c  =  0,  and  determine  the  values  of  the  coefficients  a, 
b,  and  c: 

1.  x2+4a:-5  =  0  3.  c2  =  4c-f  1 

2.  y^-2y=n  4.  a?  =  7a-l 
Change  the  following  equations  to  the  normal  form : 
6.  ax'^-\-bz  =  b-\-ax                      7.  2z'^-{-ab=^2az-\-bz 
6.  2y'^+^ay+2ab=-by  8.  s'^-\-a^  =  2(is-2 


TRIANGLES.    QUADRATICS.    RADICALS  127 

236.  Solution  of  the  equation  ax'^+bx+c=0.  Since 
every  quadratic  equation  may  be  changed  to  the. normal 
form  ax^-{-bx-{-c  =  0,  we  may  obtain  a  solution  of  every 
quadratic  equation  by  solving  ax'^-\-hx-\-c  =  0.  Thus,  we 
shall  derive  a  formula,  by  means  of  which  the  solution 
of  any  quadratic  equation  may  readily  be  found. 

Give  reasons  for  every  step  in  the  following  solution: 


ax^+hx+c 

=  0 

ax'^+hx 

=  -c 

x^+h 
a 

_     c 
a 

52 
Completing  the  square  on  the  left  side  by  adding  -^-^y 

to  both  sides  of  the  equation  we  have — 

a       4:0?     4a/-     a 

^  ,  hx  ,   b^      ¥      4ac 
a     4a?    4a^     4a^ 

^  ,  bx  ,   ¥      ¥  —  4ac 
a     4a^        4a^     • 

b^—4ac 


Whence 


'      V^2a)- 


4a2 


and  x+-=^y^- 


4ac 


Whence,  x  = 


4a2 


b  l/b^-4ac 


2a  2a 

-b=^Vb'^-4ac 
2a 


128  SECOND-YEAR  MATHEMATICS 

237.  General    quadratic    formula.     The    values    of 
X  in  the  equation — 

have  been  found  to  be 


-bW^- 

-4ac 

2a 

-b-Vb"- 

-4ac 

'"'■  2a 

These  are  the  general  quadratic  formulas.     They  may 
be  combined  into  a  single  formula  thus, 

-b^Vb'-4ac 


x= 


2a 


EXERCISES 

By  means  of  the  quadratic  formula  solve  the  following  equa- 
tions. In  these  equations  consider  o,  h,  and  c  as  knowns  and 
all  other  letters  as  unknowns: 

1.  3a;2+5x-2  =  0 
Here  a  =  3,  6  =  5,  c  =  —2. 
Substituting  these  values  in  the  formula, 

^^-5-1^25+24^-5-7^1  ^^_2. 
6  6         3' 

2.  2x2+5x+2  =  0  9.'x2+fa:=l 

3.  6a;2-llx+5  =  0  JlO.  r2-9r-36  =  0 

4.  2r2-r-6  =  0  11.  ^2+15^= -44 
6.  2x2+a;  =  15                            1:12.  a:2_72  =  6a; 

6.  1.4x2+5a;  =  2.4  JlS.  3m2  =  6-7m 

7.  I^x2+a:-11.2  =  0  14.  6+lla:= -18x2-20 

8.  .6x2-1.4x  =  3.2  16.  14?/2+2?/  =  28?/-10?/2+5 


TRIANGLES.    QUADRATICS.    RADICALS  129 


$16.  11722-10/2  =  24-10/22        :j:23.  y^+my-\-n  =  0 


17.  6p2-13p  =  10p-21 
tl8.  8^2-12^+3  =  0 

19.  s2-2as+a2+2  =  0 

20.  t''-Sabt+2a%^  =  0 

21.  a-y^=il-a)ij 

22.  cy^+ly+r=0 


24.  8?/2+8c?/+2c2=-19c2-6i/2 
$25.  2862=_l76i/+32/2 

26.  12m2-16am-3a2  =  0 

27.  ax'^+ib-a)x-b  =  0 

28.  2i/2+(4a+6)?/=-2a6 
$29.  2z^-{2a+b)z-\-ab=0 


Solve  the  following  problems : 

30.  The  diagonal  of  a  rectangle, 
Fig.  147,  is  17  inches.  One  of  the  sides  is 
7  in.  longer  than  the  other.  Find  the 
length  of  each  side. 


17. 


X^  7 

Fig.  147 


31.  The  diagonal  of  a  rectangle  is  8  units  longer  than  one  side 
and  9  units  longer  than  the  other.     How  long  is  the  diagonal  ? 

32.  A  ladder  33  ft.  long  leans  against  a  house.  The  foot 
of  the  ladder  is  14  ft.  from  the  house.  How  far  from  the 
ground  is  the  point  of  the  house  touched  by  the  top  of  the 
ladder  ? 

33.  The  diagonal  of  a  rectangle, 
Fig.  148,  is  26.  The  distance  from  the 
vertex  to  the  diagonal  is  12.  Find  the 
segments  into  which  the  perpendicular 
divides  the  diagonal. 


Fig.  148 


34.  The  height,  y,  to  which  a  ball  thrown  vertically  upward, 
with  a  velocity  of  100  ft.  per  second,  rises  in  x  seconds  is  given 
by  the  formula  y=  lOOx— 16x2.  jj^  j^q^  many  seconds  will  the 
ball  rise  to  a  height  of  144  feet  ? 

Make  a  graph  of  the  function  lOOx— 16^2  and  by  means  of 
this  graph  interpret  the  meaning  of  the  solutions  of  the 
equation. 


130  SECOND-YEAR  MATHEMATICS 

238.  Historical  note :  To  solve  a  pure  quadratic  equation, 
such  as  a:2  =  25,  is  merely  to  extract  a  square  root.  A  way  of 
extracting  square  roots  of  numbers  has  been  known  since  the 
dawn  of  history.  Early  mathematical  students  did  what 
amounted  to  solving  a  pure  quadratic  long  before  they  even 
thought  about  quadratic  equations. 

But  no  one  could  have  written  the  tenth  book  of  Euclid's 
Elements  (300  B.C.)  without  a  good  knowledge  of  ways  of  solving 
quadratic  equations.  Since  this  tenth  book  contains  most  of 
Euclid's  original  work,  it  may  safely  be  assumed  that  Euchd 
had  this  knowledge.  He  solved  no  quadratics  algebraically, 
but  he  proved  geometrical  theorems  that  amounted  to  such  solu- 
tions. Euclid  was  a  Greek  and  Greek  geometers  did  not  like 
calculatory  processes  Hke  solving  quadratics,  because  they  did 
not  think  practical  numerical  calculating  scientific  work. 
Plato  (429-348  B.C.)  had  said  calculating  is  a  childish  art  beneath 
the  dignity  of  a  philosopher. 

The  great  skill  of  Archimedes  (287-212  B.C.)  in  difficult, 
calculations,  makes  men  think  that  he  also  must  have  known 
how  to  solve  quadratics  algebraically,  but  his  writings  contain 
nothing  about  it. 

Heron  of  Alexandria  (first  century  B.C.)  was  a  scientific  engi- 
neer and  surveyor  and  he  solved  correctly  numerous  quadratic 
equations.  In  his  Geometria  he  solves  a  problem  leading  to  a 
quadratic,  which  in  modern  symbofism,  is — 

in  which  *S  is  a  given  number  and  d  is  the  diameter  of  a  circle. 
He  gives  correctly  a  rule  which  in  modern  form  is — ■ 


^     1^154^+841-29 
d=  ^^ 

Thus  by  Heron's  time  the  algebraic  rule  had  become  entirely 
dissociated  from  geometry,  and  was  known  and  studied  for  itself, 
without  any  connection  with  geometrical  theorems  of  area  or  of 


TRIANGLES.    QUADRATICS.    RADICALS  131 

lines.     It  had  taken  centuries,  however,  to  bring  about  this 

separation  from  geometry. 

The  next  important  appearance  of  the  solution  of  the 

quadratic  equation  is  in  the  Arithmetic  of  Diophantus  (third  and 

fourth  centuries  a.d.).    He  distinguishes  three  normal  forms, 

viz. — 

1.  ax^+bx  =  c  2.  ax'^  =  bx+c  3.  ax'^+c  =  bx 

As  the  Greeks  knew  no  negative  numbers,  the  three  forms 

had  to  be  kept  separate  for  treatment,  and  of  course  they  could 

not  handle  the  form — 

x'^-\-px-{-q  =  0, 

for  it  requires  a  knowledge  of  both  negative  and  complex  num- 
bers, which  neither  antiquity  nor  later  times  until  the  seven- 
teenth century  B.C.  was  able  to  comprehend. 

The  union  of  the  three  normal  forms  into  one  was  first  accom- 
phshed  by  the  Hindus.  The  rule  of  Brahmagupta  (b.  598  a.d.), 
which  was  assumed  as  known  by  his  predecessor  Aryabhatta 
(b.  476  A.D.),  expressed  in  modern  form,  was — 


ax'^-\-bx  =  c,  whence  x= 


the  agreement  of  which  with  Diophantus'  first  form  perhaps 
suggests  a  Greek  origin  of  Hindu  algebraic  knowledge. 

A  later  Hindu  scholar,  Cridhara,  introduced  a  slight  im- 
provement by  changing  the  form  to  the  following: 

_V4:ac-\-b'-b 
""'  2a 

The  eastern  Arab  Alkarchi  (about  1010  a.d.),  who  was  the 
greatest  Arabian  algebraist,  introduced  the  higher  degree  equa- 
tions of  quadratic  form — 

ax^^+bx^  =  c;  ax^^  =  &x" + c;  and  ax^^ + c  =  6x", 

and  solved  them  by  reducing  them  to  the  three  principal  cases. 


132 


SECOND-YEAR  MATHEMATICS 


Mediaeval  European  mathematicians  before  Cardan  (1501- 
76'),  still  unable  to  construe  the  significance  of  negative  number, 
continued  to  split  up  the  solution  of  quadratics  into  numerous 
special  cases,  often  including  as  many  as  24  special  cases  each 
with  its  special  rule  of  reckoning.  Finally,  Cardan  succeeded 
in  gaining  the  correct  insight  into  negative  number,  and  the 
ItaUan  school  of  chinkers  attacked  the  imaginary.  Through 
the  work  of  this  school  it  became  possible  to  supply  the  lacking 
form — 

x'^-\-pz-{-q  =  0,  for  the  cases  of  p>0  and  q>0. 


The  Generalization  of  the  Theorem  of  Pythagoras 

239.  In  the  right  triangle  ABC,  Fig.  149,  imagine  the 
angle  ABC  to  decrease,  leaving  the  lengths  of  the  sides 


B 


Fig.  149 


Fig.  150 


Fig.  151 


AB  and  BC  unchanged.  Then  the  squares  on  AB  and 
BC  are  not  changed  in  size,  but  as  the  distance  between 
the  endpoints  A  and  C,  of  AB  and  BC,  decreases,  the 
square  on  AC  decreases,  Fig.  150.  Therefore  in  a  triangle 
the  square  on  the  side  opposite  an  acute  angle  is  less  than 
the  sum  of  the  squares  on  the  other  two  sides. 

In  a  similar  way,  by  increasing  the  right  angle  ABC, 
Fig.  149,  as  in  Fig.  151,  we  find  that  the  square  on  the  side 
opposite  the  obtuse  angle  B  is  greater  than  the  sum  of  the 
squares  on  the  other  two  sides. 


TRIANGLES.    QUADRATICS.    RADICALS 


133 


The  following  two  theorems  will  show  by  how  much 
the  square  on  one  side  of  a  triangle  differs  from  the  sum 
of  the  squares  on  the  other  two  sides. 

240.  The  square  on  the  side  opposite  an  acute 
angle. 

Let  Z  5  be  an  acute  angle  of  triangle  ABC,  Fig.  152. 


Fig.  153 


Draw  CD  perpendicular  to  AB.  Denote  the  projec- 
tion of  a  on  c  by  a'. 

Then  ¥  =  h''+{c-ay.        Why? 

And  d'  =  h'-\-a'\ 

Subtracting,    ¥-a'  =  {c-a'y-a'^  =  c''-2ca'+a''-'a'\ 

Therefore        h''-a^  =  c''-  2ca\ 

Solving  for  ¥,  h^  =  a^-\-c^-2ca\ 

This  shows  that  the  product  2ca'  is  the  amount  by 
which  a^-{-c^  exceeds  ¥. 

Hence,  we  have  proved  the  following  theorem: 

Theorem:  In  a  triangle  the  square  on  the  side  opposite 
an  acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides,  diminished  by  two  times  the  product  of  one  of 
these  two  sides  and  the  projection  of  the  other  upon  it. 

EXERCISES 

1.  Find  a',  Fig.  152,  when  a,  h,  and  c  are  respectively  2,  4,  5; 
7,  10,  8. 

2.  Prove  the  theorem  in  §  240,  using  Fig.  153. 


134  SECOND-YEAR  MATHEMATICS 

241.  The  square  on  the  side  opposite  an  obtuse  angle. 

Theorem:  In  a  triangle  the  square  on  the  side  opposite 
an  obtuse  angle  is  equal  to  the  sum  of  the  squares  on  the  other 
two  sideSy  increased  by  two  times 
the  product  of  one  of  them  and  the 
projection  of  the  other  upon  it. 


Given  A  ABC  with  Z  ABC 
obtuse,  Fig.  154. 

To  prove  b^ = a^+c^-{-2ca' 


Fig.  154 


Proof: 


b'  =  ¥+(c+a'y 
a2  =  /i2-|-a'2 


Why? 
Why? 


Therefore  ¥-a^  =  c^+2ca'+a'^-a'^ 
Hence,  ¥  =  a^+c'^+2ca'. 


Why? 


EXERCISE 

The  side  opposite  an  obtuse  angle  is  6,  and  c'  is  the  pro- 
jection of  c  upon  a,  Fig.  155. 

Find  a'  and  c'  when  a,  h,  and  c  are 
respectively 

1.  5,  15,  12 

2.  6,  12,  8 
t3.    7,  11,  8 

4.    s2-l,  s2+2,  2s 

and  in  each  case  compare  2a' c  with  2ac' 


Summary 

242.  The  chapter  has  taught  the  meaning  of  the  follow- 
ing terms : 

projection  of  a  point  quadratic  formula 

projection  of  a  segment       radical 

mean  proportional  reduction  of  radical  to  simplest  form 


TRIANGLES.    QUADRATICS.    RADICALS  135 

243.  The  following  theorems  were  proved: 

I.  Theorems  expressing  relations  between  the  sides  of  a 
triangle: 

1.  In  a  right  triangle  the  square  of  the  hypotenuse  is 
equal  to  the  sum  of  the  squares  of  the  sides  of  the  right 
angle. 

2.  In  a  triangle  the  square  on  the  side  opposite  an  acute 
angle  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
diminished  by  two  times  the  product  of  one  of  these  two  sides 
and  the  projection  of  the  other  upon  it. 

3.  In  a  triangle  the  square  on  the  side  opposite  the  obtuse 
angle  is  equal  to  the  sum  of  the  squares  on  the  other  two  sides, 
increased  by  two  times  the  product  of  one  of  them  and  the 
projection  of  the  other  upon  it. 

II.  Theorems  on  mean  proportionals: 

1.  In  a  right  triangle  the  perpendicular  from  the  vertex 
of  the  right  angle  to  the  hypotenuse  is  the  mean  proportional 
between  the  seginents  of  the  hypotenuse. 

2.  In  a  right  triangle  either  side  of  the  right  angle  is  the 
mean  proportional  between  its  projection  upon  the  hypotenuse 
and  the  entire  hypotenuse. 

3.  A  perpendicular  to  a  diameter  of  a  circle  at  any  point, 
extended  to  the  circle,  is  the  mean  proportional  between  the 
segments  of  the  diameter.  . 

III.  Similarity  in  the  right  triangle: 

The  perpendicular  to  the  hypotenuse  from  the  vertex  of 
the  right  angle  divides  a  right  triangle  into  parts  similar 
to  each  other  and  to  the  given  triangle. 


136  SECOND-YEAR  MATHEMATICS 

244.  The  following  constructions  were  taught : 

1.  To  construct  a  mean  proportional  between  two 
segments. 

2.  To  construct  a  square  equal  to  a  given  rectangle. 

3.  To  construct  the  square  root  of  a  number. 

245.  Quadratic  equations  may  be  solved  by  graph, 

by  factoring,  by  completing  the   square,  and  by  the 

formula: 

-h=t=V¥-4ac 

^  —  - ^ 

2a 

where  a,  6,  c  are  the  coefficients  in  the  equation — 

246.  The  following  principle  is  useful  in  reducing 
radicals  to  the  simplest  form: 

The  square  root  of  a  product  may  be  found  by  taking 
the  square  root  of  the  factors  and  then  taking  the  product 
of  these  square  roots. 

In  symbols  the  principle  may  be  stated  thus: 

Vab  =  V~a  Vb 


CHAPTER  IX 

TRIGONOMETRIC  RATIOS.    RADICALS.     QUADRATIC 
EQUATIONS  IN  TWO  UNKNOWNS 

Trigonometric  Ratios 

247.  Finding  angles  and  distances.  The  theorem 
of  Pythagoras,  the  fact  that  two  right  triangles  are  similar 
if  an  acute  angle  of  one  equals  an  acute  angle  of  the 
other,  and  the  principle  that  the  acute  angles  of  a  right 
triangle  are  complementary,  enable  us  to  work  out  a 
method  for  finding  unknown  angles  and  distances. 

These  principles  are  the  basis  of  trigonometry,  a  sub- 
ject which  is  useful  not  only  in  the  study  of  more  advanced 
mathematics,  but  also  in  all  the  exact  sciences. 

EXERCISES 

1.  Show  that  all  right  triangles  having  an  acute  angle  of  one 
equal  to  an  acute  angle  of  the  other,  are  similar. 

2.  On  squared  paper  draw  a  right  triangle  having  an  angle 
of  30°.  Measure  the  sides  to  two  decimal  places  and  find  the 
ratio  of  the  side  opposite  the  angle  of  30°  to  the  hypotenuse. 

3.  Prove  that  this  ratio  is  the  same  for  all  right  triangles, 
having  an  angle  of  30°. 

4.  In  the  triangle  of  exercise  2,  find  approximately  to  two 
decimal  places  the  ratio  of  the  side  opposite  the  angle  60°  to  the 
hypotenuse.  Compare  your  result  with  the  results  obtained  by 
other  members  of  the  class. 

5.  Prove  that  this  ratio  is  constant  for  all  right  triangles 
that  have  an  angle  of  60°. 

6.  In  a  right  triangle  having  an  angle  of  45°,  find  the  ratio 

to  two  decimal  places  of  the  side  opposite  the  angle  45°  to  the 

hypotenuse. 

137 


138 


SECOND-YEAR  MATHEMATICS 


7.  Prove  that  this  ratio  is  constant  for  all  right  triangles 
having  an  angle  of  45°. 

8.  Draw  with  a  protractor  an 
angle  of  40°,  Fig.  156.  From 
points  on  either  side  of  the  angle 
as  Ai,A2,A3,  draw  perpendiculars 
to  the  other  side.  Measure  AiCi, 
and  AiO  and  find  their  ratio. 


Fig.  156 


9.  Prove  that  the  ratio  of  the  side  opposite  the  angle  40° 
to  the  hypotenuse  is  the  same  for  all  triangles  of  Fig.  156. 

Exercise  9  illustrates  the  fact  that  the  ratio  of  the  sides, 
Fig.  156,  remains  constant  as  the  lengths  of  the  sides  vary. 

The  constant  ratio  of  the  opposite  side  to  the  hypot^ 
enuse,  as  in  Fig.  156,  is  called  the  sine  of  angle  40°. 

248.  Trigonometric  ratios  of  an  angle.  Let  angle  A, 
Fig.  157,  be  a  given  angle.  From  any  point,  as  B,  on 
either  side  of  the  angle  draw  a  perpendicular  to  the  other 
side.     Thus,  a  right  triangle  is  formed,  as  ABC. 

In  this  triangle,  the  ratio  of  the 
side  opposite  the  vertex  of  Z  A  to  the 
hypotenuse  is  the  sine  of  angle  A* 
(written:  sin  A), 

I.e.,  sm  A=-. 
c 

*  The  word  "sine"  is  a  shortened  form  of  the  latin  siniLS,  which  is 
the  translation  of  an  Arabic  word  meaning  a  "bay,"  or  "gulf." 
Albert  Girard  (1595-1632),  a  Dutch  mathematician,  was  the  first 
to  use  the  abbreviations  "sin,"  and  "tan"  for  "sine"  and  "tangent" 
(Ball,  p.  235).  Ball  (p.  243)  says  the  term  "tangent"  was  intro- 
duced by  Thomas  Finck  (1561-1646)  in  his  Geometriae  Rotundi  of 
1583.  The  same  historian  says  (p.  243)  the  term  "cosine"  was 
first  employed  by  E.  Gunter  in  1620  in  his  Canon  on  Triangles,  and 
that  the  abbreviation  "cos"  for  "cosine"  was  introduced  by  Ough- 
tred  in  1657.  These  contractions,  "sin,"  "cos,"  and  "tan,"  did 
not  however  come  into  general  use  until  the  great  Euler  reintroduced 
them  in  1748.  The  word  "cosine"  is  an  abbreviation  for  "comple- 
mentary sine." 


RATIOS.    RADICALS.    QUADRATICS 


139 


The  ratio  of  the  side  adjacent  to  the  vertex  of  Z  A  to 
the  hypotenuse  is  the  cosine  of  angle  A  (written:  cos  A), 

I.e.,  cos  A  =  -. 
c 

The  ratio  of  the  side  opposite  to  the  side  adjacent  is 

the  tangent  of  angle  A  (written:  tan  A), 

a 


i.e.,  tan  A  = 


h' 


Stated  more  compactly: 


sin  A=the  ratio,  Typotlnufe'>  ^^  ^^e  quotient,  ^ 

cos  A=the  ratio,  ^^^|^,  or  the  quotient,  - 

tan  A=the  ratio,  "^^^^^  or  the  quotient,  ^ 

249.  Values  of  the  trigonometric  ratios  determined 
by  means  of  a  drawing.  The  values  of  the  trigonometric 
ratios  of  a  given-  angle  may  be  found  from  a  drawing  of  a 
right  triangle  containing  the  angle,  as  shown  in  the  follow- 
ing exercises : 

EXERCISES 

1.  Find  the  numerical  value 
of  sin  50°. 

With  a  protractor  construct  on 

squared  paper  an  angle  equal  to  50°, 

Fig.  158.  Draw  AB±CB.  Measure 

AB  and  AC.    Find  the  value  of  the 

AB 
ratio  -JY<.    This    is    the    required 

.  number. 

2.  Find  the  numerical  value 
of  sin  20°;  sin  45°;  sin  60°;  sin  70°. 

3.  The  values  of  the  trigono- 
metric ratios  of  angles  from  1°  Fig.  158 


1 — 

"" 

~" 

T 

" 

" 

-r 

" 

~" 

/ 

/ 

/ 

/ 

k 

- 

/ 

/ 

/ 

/ 

/ 

/ 

/ 

^ 

/ 

' 

, 

/ 

/ 

)° 

i 

/  6 

_ 

r 

f? 

140 


SECOND-YEAR  MATHEMATICS 


TABLE  OF  SINES,  COSINES,  AND  TANGENTS  OF 
ANGLES  FROM  r-90° 


Angle 

Sine 

Cosine 

Tangent 

Angle 

Sine 

Cosine 

Tangent 

r 

.0175 

.9998 

.0175 

46° 

.7193 

.6947 

1.0355 

2 

.0349 

.9994 

.0349 

47 

.7314 

.6820 

1.0724 

3 

.0523^ 

.9986 

.0524 

48 

.7431 

.6691 

1.1106 

4 

.0698 

.9976 

.0699 

49 

.7547 

.6561 

I . 1504 

6 

.0872 

.9962 

.0875 

60 

.7660 

.6428 

1.1918 

6 

.1045 

.9945 

.1051 

51 

.7771 

.6293 

1.2349 

7 

.1219 

.9925 

.1228 

52 

.7880 

.6157 

1.2799 

8 

.1392 

.9903 

.1405 

53 

.7986 

.6018 

1.3270 

9 

.1564 

.9877 

.1584 

54 

.8090 

.5878 

1.3764 

10 

.1736 

.9848 

.1763 

56 

.8192 

.5736 

1.4281 

11 

.1908 

.9816 

.1944 

56 

.8290 

.5592 

1.4826 

12 

.2079 

.9781 

.2126 

57 

.8387 

.5446 

1.5399 

13 

.2250 

.9744 

.2309 

58 

.8480 

.5299 

1.6003 

14 

.2419 

.9703 

.2493 

59 

.8572 

.5150 

1.6643 

15 

.2588 

.9659 

.2679 

60 

.8660 

.5000 

1 . 7321 

16 

.2756 

.9613 

.2867 

61 

.8746 

.4848 

1.8040 

17 

.2924 

.9563 

.3057 

62 

.8829 

.4695 

1.8807 

18 

.3090 

.9511 

.3249 

63 

.8910 

.4540 

1.9626 

19 

.3256 

.9455 

.3443 

64 

.8988 

.4384 

2.0503 

20 

.3420 

.9397 

.3640 

65 

.9063 

.4226 

2.1445 

21 

'  .3584 

.9336 

.3839 

66 

.9135 

.4067 

2.2460 

22 

.3746 

.9272 

.4040 

67 

.9205 

.3907 

2.3559 

23 

.3907 

.9205 

.4245 

68 

.9272 

.3746 

2.4751 

24 

.4067 

.9135 

.4452 

69 

.9336 

.3584 

2.6051 

26 

.4226 

.9063 

.4663 

70 

.9397 

.3420 

2.7475 

26 

.4384 

.8988 

.4877 

71 

.9455 

.3256 

2.9042 

27 

.4540 

.8910 

.5095 

72 

.9511 

.3090 

3.0777 

28 

.4695 

.8829 

.5317 

73 

.9563 

.2924 

3.2709 

29 

.4848 

.8746 

.5543 

74 

.9613 

.2756 

3.4874 

30 

.5000 

8660 

.5774 

76 

.9659 

.2588 

3.7321 

31 

.5150 

.8572 

.6009 

76 

.9703 

.2419 

4.0108 

32 

.5299 

.8480 

.6249 

77 

.9744 

.2250 

4.3315 

33 

.5446 

.8387 

.6494 

78 

.9781 

.2079 

4.7046 

34 

.5592 

.8290 

.  6745 

79 

.9816 

.1908 

5.1446 

35 

.5736 

.8192 

.7002 

80 

.9848 

.1736 

5.6713 

36 

.5878 

.P090 

.7265 

81 

.9877 

.1564 

6.3138 

37 

.6018 

.7986 

.7536 

82 

.9903 

.1392 

7.1154 

38 

.6157 

.7880 

.7813 

83 

.9925 

.1219 

8.1443 

39 

.6293 

.7771 

.8098 

84 

.9945 

.1045 

9.5144 

40 

.6428 

.7660 

.8391 

85 

.9962 

.0872 

11.4301 

41 

.6561 

.7547 

.8693 

86 

.9976 

.0698 

14.3006 

42 

.6691 

.7431 

.9004 

87 

.9986 

.0523 

19.0811 

43 

.6820 

.7314 

.9325 

88 

.9994 

.0349 

28.6363 

44 

.6947 

.7193 

.9657 

89 

.9998 

.0175 

57.2900 

45 

.7071 

.7071 

1.0000 

90 

1.0000 

.0000 

00 

RATIOS.     RADICALS.     QUADRATICS  141 

to  90°  are  tabulated  in  the  table  on  p.  140.  (Jompare  your  results 
for  exercises  1  and  2  with  the  corresponding  values  given  in 
the  table. 

250.  Values  of  the  trigonometric  ratios  found  by  means 
of  the  table.  The  table  on  p.  140  gives  approximately 
to  4  places. the  values  of  the  ratios  for  angles  containing 
an  integral  number  of  degrees  from  1°  to  90°.  This  is 
quite  sufficient  for  our  purposes. 

Where  greater  accuracy  is  required,  tables  are  avail- 
able which  give  the  values  of  the  trigonometric  ratios  of 
angles  containing  fractions  of  degrees. 

EXERCISE 

From  the  table  find  the  values  of  the  following  ratios: 
sin    2°  cos  11°  tan  20° 

sin  42°  cos  63°  tan  85° 

State  your  results  in  the  form  of  equations. 

251.  Trigonometric  functions.  Examine  the  table, 
p.  140,  and  notice  how  the  values  of  the  trigonometric 
ratios  change  as  the  angle  changes  from  1°  to  90°.  Since 
a  change  in  the  angle  produces  a  corresponding  change  in 
the  ratio,  the  trigonometric  ratios  are  also  called  trigo- 
nometric functions. 

From  the  table,  obtain  the  changes  of  sin  A  as  A  increases 
from  0°  to  90°. 

Similarly,  obtain  the  changes  of  cos  A. 

Having  given  the  value  of  a  single  function  of  an  angle 
the  values  of  the  other  functions  and  the  number  of 
degrees  in  the  angle  may  be  determined  in  various  ways. 
If  a  table  of  trigonometric  functions  is  available,  they  may 
be  looked  up  in  the  table.  An  algebraic  method  is  given 
in  §  262.     The  following  is  a  graphical  method. 


142 


SECOND-YEAR  MATHEMATICS 


252.  Graphical  method  of  finding  the  values  of  the 
functions  of  an  angle  when  one  of  them  is  known. 

EXERCISES 

1.  Given  the  sine  of  an  angle  equal  to  |-,  find  the  values  of 
the  other  functions  and  the  number  of  degrees  in  the  angle. 

Draw  a  right  angle,  A,  Fig.  159. 

On  one  side  of  the  angle  lay  off 
AB  =  l. 

With  B  as  center  and  radius  equal  to 
2  draw  a  circle  arc  meeting  AC  at  C. 

Measure  AC  and  find  the  values  of 
the  cosine  and  tangent  of  angle  C. 

With  a  protractor  find  the  number  of  degrees  in  Z  C. 

2.  Find  the  number  of  degrees  in  an  angle  whose  sine  is  |; 
.  2;    .  75.    Also  find  the  values  of  the  other  functions. 

3.  Find  the  angle  and  the  values  of  the  other  two  functions 
ifcos5  =  0.6;  iftanA=|. 

Exact  Values  of  the  Functions  of  30°,  45°,  and  60°. 

253.  Values  of  the  functions  of  30°  and  60°.  Since 
angles  of  30°,  45°,  and  60°  are  used  in  a  large  number  of 
problems,  the  student  should  remember  the  exact  values 
of  the  functions  of  these  angles,  as  found  in  the  following 
exercises: 

EXERCISES 

1.  To  construct  a  right  triangle  containing  an  angle  of  30°, 
draw  an  equilateral  triangle,  Fig.  160,  and  divide  it  into  two 
congruent  triangles  by  drawing  the  alti- 
tude to  one  side.  c 

Show  that  the  acute  angles  of  triangle 
ADC  are  60°  and  30°. 

Show  that  the  hypotenuse  is  twice  as 
long  as  the  side  opposite  the  30°—  angle. 

Hence,  if  AD  be  denoted  by  x,  AC 
must  be  2x.  _  ^       •<^     d 

Show  that  CD=xVs.  Fig.  160 


/ 

\ 

/SO 

\. 

2x/ 

^  \ 

/go° 

'JO 

\ 

RATIOS.    RADICALS.    QUADRATICS  143 

2.  Find  the  value  of  sin  30°,  using  Fig.  160. 
smSO°  =  ~=l        Why? 
.  3.  Find  the  value  of  sin  60°. 

4.  Find  the  value  of  cos  30°. 

5.  Find  the  value  of  cos  60°. 

6.  Find  the  value  of  tan  30°. 

.   ^ono     -^ 1^3    _Vs 

tan  60  =  _   /-  =    , —    ,-  =  — — - 
^V  3     VSVS       3 

7.  Find  the  value  of  tan  60°. 

254.  Rationalizing  the  denominator.     In  exercise  6, 
the  fraction  -y=  was  changed  to  ^Vs  by  multiplying 

y    O 

numerator  and  denominator  by  V^3.  This  does  not 
change  the  value  of  the  fraction  but  changes  the  de- 
nominator to  a  rational  number.  This  process  is  called 
rationalizing  the  denominator.  The  object  of  the  ration- 
alizing process  is  to  obtain  a  form  of  the  fraction  more 
easily  calculated  arithmetically. 


EXERCISES 
Rationalize  the  denominators  in  the  following  fractions: 

1  12 

1.  -7^  4.  —7-= 

VI  ^   /6-1/3 

2.  — ^  ^'  7= — 

V  2  21/3 

6  ^    V16-V2 


144  SECOND-YEAR  MATHEMATICS 

Vc  *        V^  '2-/3 

3 

To  rationalize  the  denominator  in  -7=  multiply  the  nume- 
rator and  the  denominator  by  2+Vs.    Thus, 
3               3(2+t/3)           6+3T/3, 


2-V3     (2 -1/3)  (2 +1/3)        4-3 


=  6+3V3. 


10.  -^  12.  -J-  14.  ^:i^ 

2+1/5  1/2-1  5+/2 

11.  -V      13.  -^-       15.  y:^^ 

3-V/5  3-f-2i/5  2->/3 

In  the  following  rationalize  the  denominator  and  then  find 
the  approximate  values  of  the  fractions  to  two  decimal  places: 

8-1^6  S-V2 


4-31/5  *  3-1/2    2+1/3 

t20.  Find  the  value  of  x  satisfying  the  equation 
5:^=i/3a  +  2x) 
and  express  it  as  a  fraction  with  a  rational  denominator. 

255.  Exact  values  of  the  functions  of  45°.  To  con- 
struct an  angle  of  45°,  draw  an  isosceles  right  triangle, 
Fig.  161. 

EXERCISES 

1.  In  the  isosceles  right  triangle  ABC, 
Fig.  161,  show  that  ^  =  C  =  45°. 

2.  Denoting  the  equal  sides  of  triangle 
ABC,  Fig.  161,  by  x,  show  that  AC=xV2. 

3.  Find  the  values  of  the  functions  of 
45°,  giving  all  results  with  rational 
denominators.  Fig.  161 


RATIOS.     RADICALS.     QUADRATICS 


145 


256.  Summary  of  the  exact  values  of  the  functions 
of  30°,  45°,  and  60°.  The  following  is  a  simple  device 
for  memorizing  these  values.  For  the  sake  of  symmetry, 
let  ^  be  written  in  the  form  ^\^1,  then 

sin  30°  =  Jl/l,  sin45°  =  |l/2,  sin60°  =  JV3. 

The  values  of  the  cosine  are  the  same  as  above,  but 
in  reverse  order,  thus : 

cos  60°  =  Jl/I,  cos  45°  =  Jv^2,  cos  30°  =  Jl/3 
This  may  be  conveniently  arranged  in  the  form  of  a 
table : 


^^--^^  Angle 

30°         \         45° 

60° 

Sine 

il/l      1       |V2 

K3 

Cosine 

il/3           il/2 

iv/1 

It  will  be  seen  in  §  262  that  it  is  not  necessary  to 
memorize  the  values  of  the  tangent-function  because 
they  are  easily  computed  from  a  simple  relation  existing 
between  the  trigonometric  functions.  However,  before 
making  a  study  of  these  relations,  we  shall  take  up  some 
of  the  practical  applications  of  the  functions. 


Application  of  the  Trigonometric  Fimctions 

257.  Determination  of  a  triangle.  We  know  that  all 
right  triangles  in  which  the  following  parts  are  equal, 
each  to  each,  are  congruent: 

1.  The  two  sides  including  the  right  angle. 

2.  A  side  and  one  acute  angle. 

3.  The  hypotenuse  and  one  of  the  other  sides. 


146 


SECOND-YEAR  MATHEMATICS 


In  other  words,  if  in  a  right  triangle  two  parts  in  addi- 
tion to  the  right  angle  are  given  {at  least  one  being  a  side), 
the  triangle  is  completely  determined,  and  may  be  con- 
structed from  these  given  parts.  The  unknown  parts 
may  be  computed  by  the  methods  of  scale  drawing,  or  by 
using  the  sine,  cosine,  and  tangent  of  the  angles,  as  will  be 
seen  in  the  following  exercises : 


EXERCISES 

1.  The  rope  of  a  flagpole  is  stretched  out  so 
that  it  touches  the  ground  at  a  point  20  ft.  from 
the  foot  of  the  pole,  and  makes  an  angle  of  73° 
with  the  ground.  Find  the  height  of  the  flag- 
pole. 

1.  Graphical  solution:  With  a  ruler  and  pro- 
tractor, draw  the  right  triangle,  ABC,  Fig.  162,  to 
scale.  By  measurement,  x  is  found  to  represent 
66  ft.  approximately. 

II.  Trigonometric  solution:  Using  the  tangent        Fig.  162 
of  ZA,  we  have: 

^  =  tan  73°  =  3. 2709,  from  the  table  on  p.  140. 

Therefore  a;  =  20  X  3 .  2709  =  65 .  418 

The  result,  65.418,  is  misleading,  as  it  gives  the  impression  that 
the  length  of  BC  has  been  determined  accurately  to  three  decimal 
places.  This  is  impossible  since  the  length  of  AC,  i.e.,  £0,  from 
which  65 .  418,  was  derived  by  multipHcation  had  not  been  deter- 
mined even  to  the  first  decimal  place.  Hence,  the  decimal  .418 
has  no  meaning  and  should  be  discarded.  The  length  of  BC  is 
said  to  be  65  ft.,  approximately. 

2.  A  balloon  is  anchored  to  the  ground  by  a  rope  260  ft. 
long,  making  an  angle  A  of  67°  with  the  ground.  Assuming  the 
rope  line  to  be  straight,  what  is  the  height  of  the  balloon  ? 

Use  the  sine  of  angle  A . 


. 

T 

B 

1      _ 

f 

7  ' 

7 

/ 

^     _ 

J 

/ 

f       xh 

1 

f 

t 

-M--i 

±S-" 

RATIOS.     RADICALS.    QUADRATICS  147 

3.  A  kite-string  300  ft.  long,  Fig.  163,  is  fastened 
to  a  stake  at  A.  The  distance  from  the  stake  to  a 
point  C  directly  under  the  kite  B  is  102 J  feet. 
Find  the  height  of  the  kite,  supposing  the  kite- 
string  to  be  straight. 

Find  the  angle  of  elevation  of  the  kite  from 
the  stake. 

I.  Graphical  solution:  Draw  the  right  triangle 
ABC  to  scale  and  measure  a  and  A. 

II.  Trigonometric  solution: 

A     102.5       „ 
^^^^=-300"=-^ 
From  the  table,  p.  140,  cos  72°=  .3090 
and  cos  73°  =.2924 

.*.  the  angle  of  elevation  of  the  kite  is  about  72°  or  73°. 

Since  ^  =  sin  72°=. 9511,  from  p.  140, 

therefore  a  =  300  X  .  95 1 1  =  285 . 

III.  Algebraic  solution:  The  value  of  a  may  also  be  obtained 
from  the  equation 

a  =  -/3002-102.52         Why? 

4.  A  vertical  pole,  8  ft.  long,  casts  on  level  ground  a  shadow 
9  ft.  long.    Find  the  angle  of  elevation  of  the  sun. 

Use  the  tangent  ratio. 

5.  The  angle  of  elevation  of  an  aeroplane  at  a  point  A  on 
level  ground,  is  60°.  The  point  C  on  the  ground  directly  under 
the  aeroplane  is  300  yd.  from  A.  Find  the  height  of  the  aero- 
plane. 

6.  What  is  the  angle  of  elevation  of  the  top  of  a  hill  500 1^  3  ft. 
high,  at  a  point  in  the  plain  whose  shortest  distance  from  the 
top  of  the  hill  is  1,000  feet  ? 

7.  What  is  the  angle  of  elevation  of  a  road  that  rises  1  ft. 
in  a  distance  of  50  ft.  measured  on  the  road  ? 


148  SECOND-YEAR  MATHEMATICS 

JS.  A  road  makes  an  angle  of  6°  with  the  horizontal.  How 
much  does  the  road  rise  in  a  distance  of  100  ft.  along  the  hori- 
zontal ? 

9.  On  a  tower  is  a  search-light  140  ft.  above  sea-level.  The 
beam  of  light  is  depressed  (lowered)  from  the  horizontal,  through 
an  angle  of  20°,  revealing  a  passing  boat.  How  far  is  the  boat 
from  the  base  of  the  tower? 

JlO.  A  boat  passes  a  tower  on  which  is  a  search-light  120  ft. 
above  sea-level.  Find  the  angle  through  which  the  beam  of 
light  must  be  depressed  from  the  horizontal,  so  that  it  may 
shine  directly  on  the  boat  when  the  boat  is  400  ft.  from  the  base 
of  the  tower. 

11.  From  the  top  of  a  chff  150  ft.  high,  the  angle  of  depres- 
sion of  a  boat  is  25°.  How  far  is  the  boat  from  the  top  of  the 
cUff? 

12.  When  an  aeroplane  is  directly  over  a  town  C  the  angle 
of  depression  of  town  B,  2\  miles  from  C,  is  observed  to  be  10°. 
Find  the  height  of  the  aeroplane. 

|13.  From  an  aeroplane,  at  a  height  of  600  ft.,  the  angle  of 
depression  of  another  aeroplane,  at  a  height  of  150  ft.  is  39°. 
How  far  apart  are  the  two  aeroplanes  ? 

14.  Two  persons,  1,200  ft.  apart,  observe  an  aeroplane 
directly  over  the  straight  line  from  one  to  the  other.  One 
person  finds  the  angle  of  elevation  of  the  aeroplane  to  be  35°; 
the  other,  at  the  same  time,  from  his  position,  finds  it  to  be 
55°.    Find  the  height  of  the  aeroplane. 

tl6.  On  the  top  of  a  tower  stands  a  flagstaff.  At  a  point 
A  on  level  ground,  50  ft.  from  the  base  of  the  tower,  the  angle 
of  elevation  of  the  top  of  the  flagstaff  is  35°.  At  the  same  point 
A,  the  angle  of  elevation  of  the  top  of  the  tower  is  20°.  Find  the 
length  of  the  flagstaff. 


RATIOS.    RADICALS.    QUADRATICS 


149 


16.  A  boy  wishes  to  determine  the  height  HK  of  a  factory 
chimney.  He  places  a  transit  first  at  B  and  then  at  A  and 
measures  the  angles  x  and  y.  The  transit  is  on  a  tripod  3j  ft. 
from  the  ground.  A  and  B  are  two  points  in  line  with  the 
chimney  and  50  ft.  apart.  What  is  the  height  of  the  chimney 
if  the  ground  is  level  and  if  a;  =  63°  and  u=  33  J°,  Fig.  164  ? 


Fig.  164 


In  Fig.  165^ 


w;/2  =  tan  63°  =  1.9626 
tan  33^  =.6620 


50+2 


(1) 

(2) 


(1)  and  (2)  are  simultaneous  equations  in  which  w  and  z  are 
the  unknown  numbers.     To  eliminate  w,  by  substitution,  we  have 


w;  =  1.96262  (from  (1)) 
By  substituting  (3)  in  (2), 

1.96262 


50+2 


=  .6620 


(3) 


(4) 


Find  the  value  of  z  in  (4).  Substitute  this  value  of  z  in  (3),  thus 
obtaining  the  value  of  w. 

Show  how  the  height  of  the  chimney  could  be  readily  found  by 
measuring  shadow-lengths,  without  using  angles.  One  method 
would  thus  furnish  a  check  on  the  other. 

17.  From  a  point  A  on  the  south  bank  of  a  river  flowing  due 
east  the  angle  of  elevation  of  the  top  of  a  tree  on  the  north 
side  is  45°.  At  a  point  B,  70  yd.  south  of  A,  the  angular 
elevation  is  30°,    Find  the  width  of  the  river. 


150 


SECOND-YEAR  MATHEMATICS 


18.  At  a  window  20  ft.  from  the  ground,  the  angle  of  depres- 
sion of  the  base  of  a  tower  is  15°,  and  the  angle  of  elevation  of 
the  top  of  the  tower  is  37°.    What  is  the  height  of  the  tower  ? 

tl9.  Village  B,  Fig.  166,  is  due 
north  of  village  C.  An  army  outpost 
is  located  at  a  point  A,  8  miles  due 
west  of  C.  B  bears  60°  east  of  north 
from  A.  An  areoplane  is  observed  to 
fly  from  C  to  5  in  a  quarter  of  an  hour. 
Find  the  average  horizontal  speed  of 
the  aeroplane. 

J20.  To  measure  the  width  of  a 
river  flowing  due  east,  a  man  selects  a 
point  A  from  which  a  tree  at  C,  on  the 
other  side  bears  60°  east  of  north.  He 
then  walks  east  from  A  until  he  finds 
a  point  B  from  which  C  bears  30°  west  of  north.  AB  is  found 
to  be  300  yards.     Find  the  width  of  the  river,  CH,  Fig.  167. 

Show  that  a;  =  1 50,  and  ?/  =  260. 

$21.  Two  aeroplanes  start  from  city  C  at  the  same  time. 
Aeroplane  A  flies  south  at  the  average  rate  of  15  mi.  an  hour. 
Aeroplane  B  flies  west.  At  the  end  of  f  of  an  hour,  aeroplane  B 
is  observed  to  bear  5lJ°  west  of  north  from  aeroplane  A.  How 
far  apart  are  the  aeroplanes  at  the  time  of  observation  ?  What 
is  the  average  speed  of  aeroplane  B  f 

$22.  A  balloon  is  directly  over  a  straight  road.  The  angles 
of  depression  of  two  buildings  on  the  road  are  34°  and  64°.  If 
the  buildings  are  65  yd.  apart,  how  high  is  the  balloon  ? 

J23.  From  a  lighthouse,  situated  on  a  rock,  the  angle  of 
depression  of  a  ship  is  12°,  and  from  the  top  of  the  rock  it  is  8°. 
The  height  of  the  lighthouse  above  the  rock  is  45  feet.  Find 
the  distance  of  the  ship  from  the  rock. 

J24.  From  an  aeroplane  the  angles  of  depression  of  the  top 
and  bottom  of  a  flagpole  55  ft.  high,  are  45°  and  67°,  respectively. 
Find  the  height  of  the  aeroplane, 


RATIOS.    RADICALS.    QUADRATICS 


151 


258.  Problems  on  isosceles  triangles.  Problems  on 
isosceles  triangles  may  be  solved  by  using  the  two  right 
triangles  into  which  an  altitude  line  from  the  vertex- 
angle  of  an  isosceles  triangle  divides  the  triangle. 


PROBLEMS 

1.  The  distance  from  a  cannon  to  a  straight  road  is  7  miles. 
If  the  range  of  the  cannon  is  10  mi.,  what  length  of  the  road 
is  commanded  by  the  cannon  ? 

Show  that  BAC,  Fig.  168, 
is  an  isosceles  triangle,  and  that 
AH  bisects  BC.  In  the  right 
triangle  ABH,  find  the  length 
of  BH. 

2.  The  arms  of  a  pair  of  "^--^ ^- 

compasses    are   opened    to  a  Fig.  168 

distance  of  6.25  cm.  between 
the  points.     If  the  arms  are  11.5  cm.  long,  what 
angle  do  they  form  ? 

In  the  isosceles  triangle  ABC,  Fig.  169,  draw 
the  altitude  AH. 

3.  A  pair  of  compasses  is  opened  to  an  angle 
of  50°.  What  is  the  distance  between  the  points 
if  the  arms  are  12.5  cm.  long?  ^-6.25 — 1 

Draw  the  altitude  of  the  isosceles  triangle.  p      ^  gg 

t4.  A  cannon  with  a  range  of  11  mi.  can 
shell  a  stretch  of  13  mi.  on  a  straight  road.     How  far  is  the 
cannon  from  the  road? 

J5.  A  clock  pendulum,  20  in.  long,  swings  through  an  angle 
of  6°.  Find  the  length  of  the  straight  Hne  between  the  farthest 
points  wjiich  the  lower  end  reaches. 

6.  A  clock  pendulum  is  25  in.  long.  .  Through  what  angle 
does  the  pendulum  swing  if  the  distance  between  the  farthest 
points  which  the  lower  end  reaches  is  6  inches  ? 


152 


SECOND-YEAR  MATHEMATICS 


7.  Two  firemen  are  playing  a  stream  of  water  on  the  wall 
of  a  burning  building  from  a  fire-hose  which  throws  water  120 
feet.  The  distance  on  the  ground  from  the  firemen  to  the  wall 
is  100  feet.  What  is  the  greatest  distance  on  the  wall  which  can 
be  reached  by  the  water  ? 

Relations  of  Trigonometric  Functions 

259.  Important  relations  between  the  sine,  cosine,  an  1 
tangent  of  an  angle  can  be  shown  by  simple  formulas. 


EXERCISES 

1.  Prove  that  if  A  is  any  acute  angle 
(sin^)2+(cos^)2=l 


In  Fig.  170 


Squaring  (1)  and  (2), 


sin  A=- 
c 

COS  A  =  - 


(sinA)2  =  ^ 
(cosA)2  =  ^ 


Adding  (3)  and  (4), 


(sin  A)2  +  (cosA)2  =  ^^   (5) 


Fig.  170 


*.•     a2+62  =  c2 
.-.     (sin  A)2  +  (cosA)2  =  l  (6) 

(sin  A)2  is  usually  written  sin^  A;  similarly  (cos  A)^  and  (tan  Ay 
are  written  cos^  A  and  tan^  A . 


2.  In  Fig.  170  prove  that  sin^  B+cos^  B  =  1. 

3.  Using  the  formula  sin^  x+cos^  x=l,  show  that 

sinx=  V^l  — cos^a;* 
and  cos:»=i^l— sin^  X 


(1) 

(2) 


*  We  shall  not  use  the  double  sign  before  the  radical  because 
we  have  found  no  meaning  for  a  negative  sine  or  cosine  of  angles, 


RATIOS.'  RADICALS.    QUADRATICS  153 

4.  From  Fig.  170  show  that 

,        .sinA 
tan  A  =  — —7- , 
cos  A 

J  ,       „     sin  5 

and  tan  5  = — 

cos  B 

260.  Trigonometric  identities.     The  two  fundamental 
relations 

sin2^+cos2i4  =  l  (1) 

tan^=^,  (2) 

cos  -4 

are  true  for  any  value  of  A.    They  are  therefore  called 
identities,  and  are  sometimes  written  thus, 

sin  A 


sin^  A+cos^  A  =  l;  tanA  = 


cos  A 


261.  Symbol  of  identity.  The  symbol,  = ,  is  read  is, 
or  is  identical  to. 

EXERCISES 

1.  In  Fig.  171  show  that — 

1.    sin  A  =  cos  B        2.     cos  A  =  sin  B, 

i.e.,  that  the  sine  of  an  angle  is  the 
cosine  of  the  complement  of  the  angle. 

2.  In  Fig.  171  show  that 

tan  A  =7 5, 

tan  B' 

i.e.,  the  tangent  of  an  angle  equals  the  re- 
ciprocal of  the  tangent  of  the  complement. 

262.  Given  the  value  of  one  fimction,  to  find  algebrai- 
cally the  values  of  the  others.  The  exercises  on  p.  154  show 
that  the  two  fundamental  identities  sin^  A+cos^  A  =  l, 

and  tan  A  = r  ,  may  be  used  to  find  the  values  of  two 

cos  A  "^ 

of  the  functions,  if  the  value  of  the  third  function  is  known. 


154  SECOND-YEAR  MATHEMATICS 

EXERCISES 

In  the  following  exercises  find  the  values  of  two  of  the  func- 
tions when — 


1.  tan  5  =  1 
Solution: 


tan5  =  ^  =  f.     Why?  (1) 

cos  5      *  "^  ^   ' 


and    sin2  5+cos2^  =  l.  (2) 

Equations  (1)  and  (2)  may  be  solved  as  simultaneous  equations 
in  the  two  unknowns,  sin  B  and  cos  B.  Sin  B  may  be  eliminated  by 
substitution,  as  follows: 

From  (1)                                               smB  =  j  cos  B  (3) 

Substituting  (3)  in  (2),  y^  cos^  B+cos^B  =  l  (4) 

Clearmg  (4)  of  fractions,  9  cos^  5+16  cos2  5  =  16  (5) 

25cos2  5  =  16  (6) 

cos2  5  =  4f  (7) 

cos  5  =  1  (8) 
From  equation  (3),                               sin5  =  f 

2.  cos  5  =  ^  6.  cos5=m  10.  cos5  =  |v^3 

3.  sm5=i  7.  smB  =  ^\^2  11.  tan 5  =  1^3 

4.  tan5  =  |  8.  cos5  =  Jv^2  12.  tan5  =  |i^3 

5.  sin5  =  0.5  9.  sin5  =  iT^3  13.  tan5=s 

263.  Exercises  1  to  13,  §  262,  illustrate  one  of 
the   uses   of  the  fundamental    trigonometric  relations, 

tanA= 7   and    sin^  A+cos^  J.  =  l.    The    study    of 

cos  A 

other  trigonometric  relations  is  postponed  until  we  have 

had  a  good  review  of  the  principles  of  the  operations  with 

arithmetic  and  algebraic  fractions.    These  principles  are 

reviewed  and  extended  in  chapter  x. 


RATIOS.    RADICALS.    QUADRATICS 


155 


Quadratic  Equations  in  Two  Unknowns 

264.  In  exercise  1,  §  262,  we  have  solved  the  system 
of  equations : 

sinB=f  cosB 
sin^  B+cos''B  =  l. 

In  this  system  of  equations  sin  B  and  cos  B  are  considered 
as  unknowns.  To  solve  the  system,  sin  B  was  eliminated  by 
substituting  f  cos  B  in  place  of  sin  B  in  the  second  equation. 
This  is  the  general  method  of  solving  a  system  of  equations  in 
two  unknowns,  of  which  one  is  linear  and  the  other  quadratic. 

Many  problems  lead  to  quadratic  equations  in  two 
unknowns.  The  following  problem  will  illustrate  further 
the  method  of  solution  to  be  used  in  solving  a  system  of 
equations  in  two  unknowns,  when  one  equation  is  of  the 
first  degree  and  the  other  of  the  second. 


ILLUSTRATIVE  PROBLEM 


In  the  right  triangle  ABC,  Fig.  172, 
construct  a  line  through  C  so  that  the 
perimeters   of  the  two  new  triangles 
formed  may  be  equal. 

Analysis:  Consider  the  problem 
solved  and  let  CD  be  the  required  line 
through  C.  The  position  of  D  evidently 
is  determined  by  determining  AD. 

Solution:  Denoting  the  length  of  AD 
by  X,  and  the  length  of  DB  by  y, 

S+x+CD  =  4:-\-y+CD. 

Hence, x—y  =  l 

AJ52=(x-h2/)2  =  32+42  =  25 
Hence,  x'^+2xy+y^  =  25 


with  sides  3  and  4,  to 


Why? 
Why? 
Why? 


(1) 

(2) 


The  values  of  x  and  y  are  the  solutions  of  the  system  of  equations 
(1)  and  (2). 


156  SECOND-YEAR  MATHEMATICS 

Solving  (1)  for  a;  and  substituting  in  (2), 

(1 +yy +2(1 +y)y+y' =  25  (3) 

y'+y-Q=0 

^^=2       and^^^=-^ 


1^1  =  3  ia;2=— 2 

The  values  x=—2,  y=—S  satisfy  equations  (1)  and  (2)  but 
do  not  satisfy  the  conditions  of  the  problem.  Therefore  this 
solution  is  disregarded  and  3  and  2  are  the  required  values  of 
X  and  y,  respectively. 

From  the  preceding  solution  it  is  seen  that  a  system  of 
equations  in  two  unknowns,  when  one  of  the  equations  is  of  the 
first  and  the  other  of  the  second  degree,  may  be  solved  as  follows : 

Solve  the  linear  equation  for  one  of  the  unknowns,  x  or  y,  and 
substitute  that  value  in  the  second-degree  equation.  This  will  lead 
to  a  second-degree  equation  in  the  other  unknown,  y  or  x,  as  (3), 
which  is  then  to  be  solved. 

The  values  of  y  or  x,  thus  found,  may  then  be  substituted  in 
the  first-degree  equation,  as  (1),  to  determine  the  corresponding 
values  of  the  other  unknown. 

Notice  that  the  method  of  solving  the  system  of  equations 
above  is  the  method  of  elimination  by  substitution. 

EXERCISES 

1.  Construct  a  right  triangle  whose  perimeter  is  30  and  whose 
hj^otenuse  is  13. 

2.  In  the  right  triangle  ABC,  Fig.  173,  the  perimeters 
of  ACD  and  BCD  are  equal.  OT  =  4,  and  DB  =  2.  Find  AC 
and  AD. 

c 


3.  In  the  right  triangle  of  Fig.  174,  with  sides  5x,  Sy,  and 
13,  x-{-y  =  5.    Construct  the  triangle. 


RATIOS.    RADICALS.    QUADRATICS  157 


4.  Solve^'+^^f  ^ 

[y-x=l 


5.  Sclve 


r+2s  =  7 


6.  Solver 

[m—n  =  S 


\m-\-n  =  7 


Quadratic  Equations  Solved  by  the  Graph 

265.  Problems  which  lead  to  quadratic  equations  in 
two  unknowns  may  be  solved  by  means  of  the  graph,  as 
follows : 

EXERCISES 

1.  Solve  a:2+7/2  =  25  and  y—x  =  l  by  the  graph. 

By  assuming  values  for  x,  and  solving  x^+y^=2^  for  y,  we  have 
the  following  solutions  of  the  equation  x^+y^  =  2b: 

•  rx  =  0        ('x  =  3        ra;  =  4        (x  =  b{x=-^    rx=-4    /x=-5 
\y  =  d=5    \i/=±4   l2/  =  =±=3   \y  =  0  W  =  ±4    l2/  =  ±3    \?/=     0 

Plotting  these  solutions,  Fig.  175,  we  find  that  the  graph  of 
-c2_j_^2  =  25  is  a  circle  whose  cen- 
ter is  at  thoyrigin  and  whose 
radius  is  1^25,  or  5. 

That  a:2+2/2  =  25  is  a  circle 
may  be  shown  as  follows: 

The  equation  expresses  the 
fact  that  the  sum  of  the 
squares  of  the  co-ordinates  of 
any  point  on  the  graph  of  the 
equation  is  25,  e.g., 

OPi2=a:i2+2/i2  =  25 

Hence,  0Pi  =  5. 

Moreover,  a  line  every 
point  of  which  has  the  same  dis-  FiQ-  175 

tance  from  a  given  point  is  a 
circle.    Therefore  the  graph  of  x'^-{-y'^  =  25  is  a  circle  whose  radius  is  5 . 

The  points  of  intersection  of  this  circle  and  the  straight-line 
graph  of  equation  y—x  =  l,  are  Pi (3,  4),  and  P2(— 4,  —3). 

3      ix=-4: 


1 

V 

^ 

..-^ 

"-^ 

\ 

/. 

v« 

,k) 

/ 

/\ 

/ 

/ 

1/r 

\ 

- 

/ 

/ 

/ 

p 

/ 

x^ 

\_l 

/ 

/ 

-f 

f(- 

\ 

/ 

/ 

\ 

/ 

"7 

V 

>^ 

-^ 

y 

\y 


y=-Z 


are  the  required  values  of  x  and  y. 


158 


SECOND-YEAR  MATHEMATICS 


12 


2.  In    triangle    ABC,    Fig.    176,    AB  =  5,    CD  =  ~  ,   angle 
C  =  90°     Construct  the  triangle. 


X 

Fig.  177 


3.  The  perimeter  of  the  rectangle,  Fig.  177  is  34.     Find  the 
dimensions. 

4.  Solve  by  eliminating  by  substitution,   and   verify  by 
graphing: 

\y=10-x  / /y 


x-dij  =  0 

5.  In  triangle  ABC,  Fig.  178,  draw 
DE  parallel  to  AB  so  that  DE  is  the 
mean  proportional  between  AC  and  DC. 


6.  Solve  the  following  systems: 

xy=lS 
x-2y  =  0 


2. 


3. 


t4. 


5. 


Sxy-5y-^l  =  0 
x-2y  =  0 

x'^-hxy-{-7/  =  7 
x-\-^y=  —1 

rx2+4?/  =  32 
\5a;+6?/  =  8 

2r2— rs  =  6s 

r+2s  =  7 


0:2-1/2  =  25 
x— ?/  =  10 


fa;2-2/2  =  9 
^^'  U+2/  =  9 


J9. 


x2+?/2  =  25 

X  +  2/=l 


^       \3a2-762  =  5 


RATIOS.     RADICALS.     QUADRATICS  159 

Summary 

266.  The  trigonometric  ratios  sine,  cosine,  and  tangent 
have  been  defined. 

267.  The  value  of  a  trigonometric  ratio  of  a  given 
angle  may  be  found  (1)  from  the  table,  (2)  graphically. 

268.  The  exact  values  of  the  sine  of  angles  of  30°, 
45°,  and  60°  are  |,  ^^2,  and  ^V  3  respectively. 

The  exact  values  of  the  cosine  of  the  same  angles  are 
J1/3,  Jv^2,  and  J,  respectively. 

The  value  of  the  tangent  is  found  from  the  relation 

.     sin  A 

tan  A  = 7  . 

cos  A 

269.  Many  problems  in  distances,  which  may  be 
solved  graphically,  can  be  solved  simply  by  calculating 
by  the  aid  of  trigonometric  functions. 

270.  The  following  fundamental  trigonometric  iden- 
tities have  been  proved: 

sin^  A+cos^  A  =  l 

.     sin  A 

tanA= -r 

cos  A 

271.  If  the  value  of  one  function  is  given  the  values 
of  the  other  functions  may  be  found,  (1)  from  the  table, 
(2)  graphically,  (3)  algebraically,  using  the  identities  in 
§270. 

272.  A  system  of  equations  in  two  unknowns,  when 
one  equation  is  of  the  first  degree  and  the  other  of  the 
second,  may  be  solved  by  the  method  of  elimination  by 
substitution. 

273.  The  irrational  denominator  of  a  fraction  may 
be  rationalized  by  ;iiultiplying  the  numerator  and  the 
denominator  by  the  same  number. 


CHAPTER  X 


THE  CIRCLE 
Review  and  Extension  of  the  Properties  of  the  Circle 

274.  Gothic  arch.  One  of  the  uses  of  the  circle  in 
designs  is  illustrated  in  Fig.  179.  It 
represents  the  so-called  equilateral 
Gothic  arch,  frequently  found  in 
modern  architecture.  Its  most 
common  use  is  in  church  windows. 
A  5  C  is  an  equilateral  triangle  and 
arcs  A  C  and  C5  are  drawn  with  centers 
Fig.  179  at  A  and  B,  respectively,  and  radius  A  B. 


EXERCISES 

1.  In  Fig.  180  three  Gothic  arches 
are  joined  with  a  circle.  Construct  this 
figure  with  ruler  and  compass. 

To  find  the  center  of  circle  0  use  A 
and  B  as  centers  and  radius  equal  to 
jAB.  In  exercise  3,  §  289,  we  shall  learn 
to  prove  that  the  circles  in  this  figure  are 
tangent  to  each  other  in  pairs. 


Fig.  181 

2.   Study  the  designs  in  Fig.  181  and  construct  them,  using 
ruler  and  compass. 

160 


DRYBUKGH  ABBEY— CLOISTER  DOOR 


Courtesy  of  Walter  Sargent 
GOTHIC  DOOR  AND  WINDOW 


THE  CIRCLE  161 

3.  Compare  the  distances  from  the  center  of  a  circle  to 
several  points  taken  anyivhere  -within,  upon,  or  outside  of,  the 
circle  with  the  length  of  the  radius. 

Exercise  3  shows  that  a  point  is  within,  upon,  or  without  a  circle 
according  as  its  distance  from  the  center  is  less  than,  equal  to,  or  greater 
than,  the  radius. 

275.  Concentric  circles.  Draw  several  circles  having 
the  same  center  but  unequal  radii.  Circles  having  the 
same  center  are  called  concentric  circles. 

EXERCISE 

On  notebook  paper  draw  two  circles  having  equal  radii.  If 
one  of  the  circles  is  cut  out  and  laid  upon  the  other  making  the 
centers  coincide,  the  circles  should  coincide.  See  if  you  can  make 
one  of  your  circles  coincide  with  the  other. 

If  they  do  not  coincide,  what  seems  to  cause  the  failure  of 
coincidence  ? 

In  general,  two  circles  having  equal  radii  are  equal,  and  equal 
circles  have  equal  radii. 

276.  Semicircle.  Major  arc.  Minor  arc.  Cut  a 
circle  from  paper.  Fold  it  along  a  diameter.  How  do  the 
two  parts  of  the  circle  compare  as  to  size? 

This  shows  that  a  diameter  divides  a  circle  into  two 
equal  parts.*  Each  of  these  parts  is  called  a  semicircle. 
If  a  circle  is  divided  into  unequal  parts,  one  is  called  the 
major  arc,  the  other  the  minor  arc. 

277.  Secant.     Tangent.     Draw 
a  circle  as  A,  Fig.  182.    Move  a 
ruler  B  across  the  circle  and  notice 
the  different  positions  of  the  edge,     '  Yiq.  182 
as  B,  Bi,  B2,  etc.    How  many  points 

may   a    circle    and    a  straight  line   have   in  common? 

*  According  to  Proclus  this  theorem  is  a  discovery  of  Thales. 


162  SECOND-YEAR  MATHEMATICS 

A  straight  line  intersecting  a  circle  in  two  points  is  a 
secant. 

A  line  touching  a  circle  in  only  one  point  is  a  tangent. 

278.  Number  of  points  common  to  two  circles.     By 
moving  one  circle  over  another,  Fig.  183,  show  that  two 


Fig.  183 

circles  must  intersect  in  two  points,  or  touch  in  one  point, 
or  have  no  point  in  common. 

279.  Chord.    A    segment   joining    two 
points  of  a  circle  is  a  chord,  Fig.   184. 

280.  Symbol  for  arc.     The  symbol  "^ 
means  arc.    Thus,  arc  AB  may  be  written        ^  o'ira 
AB, 

Draw  two  equal  circles.  Lay  one  circle  upon  the 
other,  making  the  centers  coincide.     If  AB  on  one  circle 

is  equal  to  CD  on  the  other,  they  can  be  made  to  coincide. 
How  do  the  chords  AB  and  CD  compare  ? 
In  a  given  circle  construct  two  equal  arcs. 

281.  Theorem:  In  the  same  or  equal  circles  equal 
central  angles  intercept  equal  arcs,  and  equal  arcs  are  inter- 
cepted by  equal  central  angles. 

For  if  the  arcs  are  made"  to  coincide  the  central  angles  coin- 
cide, and  conversely. 

282.  Subtending  chord.  The  chord  joining  the  end- 
points  of  an  arc  subtends  (stretches  under  or  across) 
the  arc. 


THE  CIRCLE 


163 


283.  Theorem:  In  the  same  or  equal  circles  equal  arcs 
are  subtended  by  equal  chords;  and  conversely,  equal  chords 
subtend  equal  arcs. 

The  truth  of  the  theorem  is 
easily  shown  by  the  method  of 
superposition. 

To  prove  the  converse, 
draw  CA,  CB,  C'A\  and  C'B\ 
Fig.  185. 

Prove     AABC^AA'B'C. 
Then,  AC^AC        Why? 

AB  =  Fb'        Why? 


Diameters,  Chords,  and  Arcs 

284.  Theorem:  A  line  drawn  through  the  center  of  a 
circle  'perpendicular  to  a  chord,  bisects  the  chord  and  the 
arcs  subtended  by  the  chord. 

Given  OO*  and  CD  drawn  through  the  center  0, 
intersecting  the  chord  AB  at  E;  also  CD^_AB,  Fig.  186. 

To  prove   AE=EB;   AD=DB; 
AC  =  CB. 

Proof   (method  of  congruent  tri- 
angles) : 

Draw  AO  and  OB, 

Prove    AAEO^ABEO 

Hence,         AE  =  EB      Why? 
and  y  =  y'         Why  ? 

Show  that  Ab  =  DB 

and  AC  =  CB 

*  The  symbol  O  0  means  the  circle  whose  center  is  0.  The 
symbol  (s)  means  circles. 


164  SECOND-YEAR  MATHEMATICS 

The  theorem  above  is  one  of  a  group  of  theorems  in- 
volving the  following  conditions: 

1.  A  line  passes  through  the  center. 

2.  A  line  is  perpendicular  to  a  chord. 

3.  A  chord  is  bisected  by  a  line. 

4.  A  minor  arc  is  bisected. 

5.  A  major  arc  is  bisected.* 

By  taking  as  hypothesis  any  two  of  these  five  conditions 
and  as  conclusion  one  of  the  remaining  three  we  can  form 
a  number  of  theorems.  Some  of  these  are  stated  among 
the  following  exercises: 

EXERCISES 

1.  A  diameter  that  bisects  a  chord  is  perpendicular  to  the 
chord  and  bisects  the  subtended  arcs.    Prove. 

2.  A  line  bisecting  a  chord  and  one  of  the  subtended  arcs 
passes  through  the  center,  is  perpendicular  to  the  chord,  and 
bisects  the  other  subtended  arc.    Prove. 

Prove  AACE^  ABCE,  Fig.  187. 

Then  CD±AB.        Why? 

Hence,  CD  passes  through  0.  For  the 
perpendicular  bisector  of  a  segment  contains 
all  points  equidistant  from  its  endpoints 
(§71). 

Show  that  AD  =  i)5 

.-.     AD^DB        Why? 

3.  The  line-segment  joining  the  mid- 
points of  the  arcs  into  which  a  chord  divides  a  circle  is  a 
diameter,  bisects  the  chord,  and  is  perpendicular  to  the  chord. 
Prove. 

J4.  A  diameter  bisecting  an  arc  is  the  perpendicular  bi- 
sector of  the  chord  subtending  the  arc.    Prove. 


THE  CIRCLE  165 

5.  The  perpendicular  bisector  of  a  chord  passes  through  the 
center  of  the  circle  and  bisects  the  subtended  arcs.     Prove. 

J 6.  A  line  perpendicular  to  a  chord  and  bisecting  one  of  the 
subtended  arcs  passes  through  the  center  of  the  circle,  and  bisects 
the  chord  and  the  other  subtended  arc.    Prove. 

7.  A  diameter  that  bisects  a  chord  bisects  the  central  angle 
between  the  radii  drawn  to  the  endpoints  of  the  chord. 

8.  Bisect  a  given  arc. 

9.  Given  a  circle,  find  the  center. 

10.  Given  an  arc,  find  the  center  and  draw  the  circle. 

11.  Draw  a  circle  through  three  points  not  lying  in  the  same 
straight  Une. 

12.  Show  that  the  perpendicular  bisectors  of  the  sides  of  an 
inscribed  polygon  meet  in  a  conamon  point. 

13.  Circumscribe  a  circle  about  a  triangle. 

|14.  Through  a  point  within  a  circle  draw  a  chord  that  will 
be  bisected  by  the  point. 

15.  Draw  a  circle  that  will  pass  through  two  given  points 
and  have  a  given  radius. 

16.  If  a  circle  is  divided  into  3  equal  parts,  and  the  points 
of  division  are  joined  by  chords,  an  equilateral  triangle  is  formed. 
Prove. 

17.  If  the  endpoints  of  a  pair  of  perpendicular  diameters 
of  a  circle  are  joined  consecutively,  what  kind  of  polygon  is 
formed  ?    Prove. 

JlS.  Show  that  the  perpendicular  to  a  tangent  at  the 
contact-point  passes  through  the  center  of  the  circle. 

19.  Construct  a  tangent  to  a  circle  at  a  given  point  of  the 
circle. 

t20.  To  a  given  circle  draw  a  tangent  that  shall  be  parallel 
to  a  given  line. 


166  SECOND-YEAR  MATHEMATICS 

285.  Theorem:  In  the  same,  or  in  equal  circles,  equal 
chords  are  equally  distant  from  the  center;  and,  conversely, 
chords  equally  distant  from  the  center  are  equal. 


Given   00=00',  AB  =  A'B'  =  A"B'' 

OC±AB,  OV'±A'B',  0C''1.A"B",  Fig.  188. 

To  prove  0C  =  0C"  =  0'C'. 

Proof  (method  of  congruent  triangles) : 

Draw  OA,  OA''  and  0'A\ 

Prove  that  A0=A'0'=A"O. 

Prove    AAOC^AA"OC"^AA'0'C\ 

Then,  OC  =  OC",  and  OC  =  O'C. 

Conversely,  If  OC"  =  0'C'  =  OC,  prove  that 

AB  =  AJB'  =  A^". 
Prove    AOAC^  AOA"C'^0'A'C'. 
Then,  AC  =  A'C  =  A"C"  2.neiAB  =  A'B'  =  A"B". 

EXERCISES 

1.  If  two  intersecting  chords  make  equal  angles  with  the  Hne 
joining  their  common  point  to  the  center,  the  chords  are  equal. 
Prove. 

2.  In  a  circle  the  distances  from  the  center  to  two  equal 
chords  are  denoted  by — 

1.  a;2+3a:  and  4(15-a;)  t3.  x{x-Z)  and  4(3x-9) 

2.  x{x+^)  and  3(2x+5)  J4.  3a;2+4rc  and  12(1 -a;) 
Find  X  and  the  distances  from  the  center  to  the  chords. 


THE  CIRCLE 


167 


286.  Theorem:  The  arcs  included  between  two  parallel 
secants  are  equal;  and,  conversely,  if  two  secants  include  equal 
arcs,  and  do  not  intersect  within  the  circle,  they  are  parallel. 

I.  Given  circle  0  and  AB  \\  CD,  cutting  the  circle  at 
A  and  B,  and  at  C  and  D,  respectively,  Fig.  189, 

To  prove  AC  =  BD. 

Proof:   Draw  OE±AB,  and  prolong 
it  to  meet  CD. 


Then, 


OE±CD. 
CE  =  DE 
AE  =  EB 
AC  =  BD 


Why? 
Why? 
Why? 
Why? 


II.  Conversely,  given  AC  =  BD,  Fig.   189,  AB  and 
CD  not  intersecting  within  the  circle, 

To  prove  AB  11  CD. 

Proof:    Draw  OE±AB  and  prolong  it  to  F. 
Prove  EC  =  ED.      Why? 

Then,  EFLCD        Why? 

.-.     ABWCD        Why? 

III.  Prove  the  theorem  with  one  of  the  lines,  as  ABj 
tangent  to  the  circle,  as  in  Fig.  190. 

G  .  E^ 


IV.  Prove  the  theorem  with  both  parallels  tangent 
to  the  circle,  as  in  Fig.  191. 

Draw  HK  \\  AB  and  apply  Case  III. 


168 


SECOND-YEAR  MATHEMATICS 


287.  Theorem:     The  line  joining  the  centers  of  two 
intersecting  circles  bisects  the  common  chord  perpendicularly. 


Fig.  192 


Fig.  193 


Let  0  and  0'  be  the  intersecting  circles,  Figs.  192, 
193.     Let  A  5  be  the  common  chord. 

To  prove  00'±AB. 

To  prove  this  apply  §  39. 

Tangent  Circles 

288.  Tangent  circles.  Two  circles  are  said  to  be 
tangent  to  each  other  if  both  are  tangent  to  the  same  line 
at  the  same  point.  This  point  is  the  point  of  tangency,  or 
the  point  of  contact  of  the  circles. 

If  the  tangent  circles  lie  wholly  without  each  other 
they  are  tangent  externally,  Fig.  194. 


Fig.  194 


Fig.  195 


If  one  of  the  tangent  circles  lies  within  the  other  they 
are  tangent  internally,  Fig.  195. 


THE  CIRCLE 


169 


289.  Theorem:   If  two  circles  are  tangent  to  each  other y 
the  centers  and  the  point  of  tangency  lie  in  a  straight  line. 


Fig.  197 


I.  Let  O  and  0'  be  the  centers  of  two  circles  tangent 
externally,  T  being  the  point  of  tangency,  Fig.  196. 

To  prove    0,  0',  and  T  lie  in  a  straight  line. 

Prove  that  OTO'  is  a  straight  angle. 

Then  OT  and  O'T  are  in  a  straight  line.    -Why  ? 

II.  Prove  the  theorem  for  the  case  shown  in  Fig.  197. 


EXERCISES 

1.  Draw  a  circle  tangent  to  a  given  circle  at  a  given  point. 
How  many  such  circles  can  be  drawn  ? 

2.  Draw  a  circle  through  a  given  point  and  tangent  to  a  given 
circle. 

3.  If  the  distance  between  the  centers  of  two  circles  is  equal 
to  the  sum  of  their  radii  the  circles  are  tangent  externally. 
Prove. 

4.  The  distance  between  the  centers  of  two  tangent  circles 
is  2  J  inches.    The  radius  of  one  is  f  inch.     Draw  the  two  circles. 

t5.  The  radii  of  three  circles  are  1  in.,  ij  in.,  and  f  in., 
respectively.    Draw  the  circles  tangent  to  each  other  externally. 

6.  Construct  a  circle  with  a  given  point  as  center  and  tangent 
to  a  given  circle. 


170  SECOND-YEAR  MATHEMATICS 

7.  To  construct  a  circle  having  a  given  radius  and  tangent 
to  two  given  circles. 

t8.  With  the  vertices  of  a 
triangle  as  centers  construct  three 
circles  tangent  to  each  other. 
(See  Fig.  198.) 

Show  algebraically  that  one  of 
the  radii  is  equal  to  half  the 
perimeter  diminished  by  one  of 
the  sides.  Fig.  198 

290.  Historical  note :  The  part  of  the  theory  of  the  circle 
that  deals  with  chords,  tangents,  and  secants  is  older  than  the 
time  of  EucHd.  Most  of  it  was  probably  first  worked  out  by 
the  Pythagoreans.  It  is  well  known  that  Archytas  of  Tarentum 
(430-365  B.C.)  at  a  certain  point  in  his  construction  of  the  prob- 
lem of  doubling  a  cube,  assumed  a  knowledge  of  the  theorem  that 
the  angle  between  a  tangent  and  the  contact-radius  is  a  right 
angle.  The  first  use  of  the  theorem  of  the  equality  of  the  two 
tangents  to  a  circle  from  an  outside  point  of  which  we  have 
knowledge  is  with  Archimedes  (287-212  b.c).  Heron  (first 
century  B.C.)  is  the  first  to  give  it  place  as  an  independent 
theorem. 

The  converse  theorem,  that  the  center  of  the  circle  lies 
on  the  bisector  of  the  angle  between  two  tangents  is  first  met 
with  in  the  seventh  book  of  the  Synagoge  of  Pappus  about 
the  end  of  the  third  century  a.d.  Archimedes  is  said  to  have 
written  an  entire  work  on  the  tangency  of  circles.  The  so-called 
tadion-problem  of  Apollonius  was  to  draw  a  circle  which  should 
fulfil  three  conditions,  viz.,  go  through  a  given  point,  be  tangent 
to  a  given  straight  line,  and  a  given  circle.  In  the  fourth  book  of 
his  Synagoge  Pappus  studied  the  problem  to  draw  a  circle  tangent 
externally  to  three  given  circles  and  treated  another  interesting 
problem  "through  three  points  of  a  straight  line  to  draw  three 
other  straight  lines  that  should  form  an  inscribed  triangle  within 
a  given  circle."  This  problem  has  more  recently  given  rise  to 
varied  generalizations. 


THE  CIRCLE  -  171 

Summary 

291.  The  meaning  of  the  following  terms  was  taught: 

concentric  circles  secant 

arc  tangent 

semicircle  chord 

major  arc  subtending  chord 

minor  arc  tangent  circles 

The  following  symbols  were  introduced :  for  chord, 
'^  for  arc,  O  for  circle,  (s)  for  circles. 

292.  The  truth  of  the  following  theorems  has  been 
shown : 

1.  A  point  is  mithin,  upon,  or  without,  a  circle  according 
as  its  distance  from  the  center  is  less  than,  equal  to,  or  greater 
than,  the  radius. 

2.  Circles  having  equal  radii  are  equal,  and  equal  circles 
have  equal  radii. 

3.  A  diameter  divides  a  circle  into  equal  parts. 

4.  In  the  same  or  equal  circles  equal  central  angles 
intercept  equal  arcs,  and  equal  arcs  are  intercepted  by  equal 
central  angles. 

293.  The  following  theorems  have  been  proved: 

1.  In  the  same  or  equal  circles  equal  arcs  are  subtended  by 
equal  chords;  atid,  conversely,  equal  chords  subtend  equal  arcs. 

2.  If  any  two  of  the  following  conditions  are  taken  as 
hypothesis  the  remaining  three  are  true: 

(1)  A  line  passes  through,  the  center. 

(2)  A  line  is  perpendicular  to  a  chord. 

(3)  A  chord  is  bisected  by  a  line. 

(4)  A  minor  arc  is  bisected. 

(5)  A  major  arc  is  bisected. 


172  SECOND-YEAR  MATHEMATICS 

3.  1 71  the  same  or  equal  circles  equal  chords  are  equally 
distant  from  the  center;  and,  conversely,  chords  equally 
distant  from  the  center  are  equal. 

4.  The  arcs  included  between  two  parallel  secants  are 
equal;  and,  conversely,  if  two  secants  include  equal  arcs, 
and  do  not  intersect  within  the  circle,  they  are  parallel. 

5.  The  line  joining  the  centers  of  two  intersecting  circles 
bisects  the  common  chord  perpendicularly. 

6.  If  two  circles  are  tangent  to  each  other,  the  centers 
and  the  point  of  tangency  lie  in  a  straight  line. 

7.  Two  arcs  are  equal  if  one  of  the  following  conditions 
holds: 

(1)  The  subtending  chords  are  diameters. 

(2)  The  central  angles  intercepting  the  arcs  are 
equal. 

(3)  The  subtending  chords  are  equal. 

(4)  The  arcs  are  intercepted  by  parallel  chords, 
secants,  and  tangents. 

8.  Two  chords  are  equal  if  one  of  the  following  condi- 
tions holds : 

(1)  The  chords  subtend  equal  central  angles. 

(2)  The  chords  subtend  equal  arcs. 

(3)  The  chords  are  equally  distant  from  the  center. 


CHAPTER  XI 
MEASUREMENT  OF  ANGLES  BY  ARCS  OF  THE  CIRCLE 

294.  Units  of  angular  measure.  In  all  preceding 
work  angles  have  been  measured  by  comparing  them  with 
such  angular  units  as  degree,  minute,  second,  right  angle, 
and  straight  angle.  Thus,  the  measure  of  an  angle  is  45, 
if  it  contains  45  degrees;  the  measure  of  the  same  angle  is 
^,  if  the  right  angle  is  used  as  unit;  or  it  is  J,  if  the  straight 
angle  is  the  unit  of  measure. 

In  the  following  it  will  be  shown  that,  if  the  sides  of  an 
angle  touch  or  intersect  a  circle,  it  is  possible  to  measure  the 
angle  in  terms  of  the  arcs  intercepted*  by  the  sides  of  the  angle. 

EXERCISES 

1.  From  cardboard  cut  a  right  angle.  Move  it  so  that  the 
sides  always  pass  through  two  fixed  points,  as  A  and  B,  Fig. 
199.  This  may  be  done  by  letting 
the  sides  always  touch  two  pins 
stuck  into  the  paper  at  A  and  B. 
Mark  the  position  of  the  vertex  for 
various  positions  of  the  angle.  How 
does  the  vertex  move  ? 

2.  Repeat   exercise    1   with  an  tjv     iqq 
acute  angle;  with  an  obtuse  angle. 

3.  Draw  a  semicircle.  Join  various  points  of  the  semicircle 
to  the  endpoints  of  the  diameter  forming  angles  whose  vertices 
lie  on  the  circle.  With  a  protractor  measure  these  angles.  How 
do  they  compare  in  size  ? 

*  Note  the  difference  between  the  words  "intercept"  and  "inter- 
sect." The  former  means- "to  hold  between"  and  the  latter  "to 
cut,  or  to  cross." 

173 


174 


SECOND-YEAR  MATHEMATICS 


4.  Draw  a  circle.  With  a  chord  cut  off  an  arc  greater  than  a 
semicircle  and  join  various  points  of  the  arc  to  the  endpoints 
of  the  chord.  By  measuring,  compare  the  angles  having  the 
vertices  on  the  arc. 

5.  Repeat  exercise  4,  using  an  arc  less  than  a  semicircle. 

295.  Inscribed  angle.  An  angle  whose  vertex  is  on  a 
circle  and  whose  sides  are  chords  is  an  inscribed  angle. 


EXEECISES 

1.  Exercises  4  and  5,  §  294,  illustrate  the  fact  that  all  in- 
scribed angles  intercepting  the  same  arc  are  equal,  and  that  the 
angle  is  acute,  right,  or  obtuse  according  as  the  intercepted  arc 
is  less  than,  equal  to,  or  greater  than,  a  semicircle. 

How  does  an  inscribed  angle  vary  as  the  arc  increases  from  a 
short  length  to  the  length  of  the  circle  ? 

2.  Show  how  a  carpenter's  square  may  be  used  to  test  the 
accuracy  of  a  semicircular  groove.     (See 

Fig.  200.) 


Fig.  200 


Fig.  201 


3.  Show  how  a  carpenter's  square  may  be  used  to  find  where 
a  ring  must  be  cut  so  that  the  two 

parts  are  equal.     (See  Fig.  201.) 

4.  The  circle  in  Fig.  202  represents 
a  Tegion  of  dangerous  rocks  to  be  avoided 
by  ships  passing  near  the  coast  AB. 
Outside  of  the  circle  there  is  no  danger. 
Show  that  the  ship  S  is  out  of  danger 
as  long  as  angle  A  SB,  found  by  observa- 
tions made  from  the  ship,  is  less  than  the 
known  angle  ACB.  Fig.  202 


MEASUREMEI^TT  OF  ANGLES  BY  ARCS 


175 


296.  If  two  lines  intersect  and  also  cut  or  touch  a 
circle,  the  various  positions  may  be  illustrated  as  in 
Figs.  203-209. 


Fig.  203 


Fig.  204 


Fig.  205 


Fig.  206 


Pig.  207 


Fig.  208 


Fig.  209 


In  Fig.  203  the  lines  intersect  at  the  center  of  the 
circle,  i.e.,  the  angle  is  formed  by  two  radii. 

In  Fig.  204  the  lines  intersect  within  the  circle,  not 
at  the  center,  i.e.,  the  angle  is  formed  by  two  chords. 

Moving  the  intersecting  lines  until  the  vertex  of  the 
angle  is  on  the  circle,  the  angle  becomes  an  inscribed  angles 
Fig.  205. 

Leaving  one  side  of  the  angle,  Fig.  205,  fixed  and  turn- 
ing the  other  until  it  is  tangent  to  the  circle.  Fig.  206  is 
obtained.  In  this  figure  the  angle  is  formed  by  a  tangent 
and  a  chord. 

Fig.  207  shows  the  lines  intersecting  outside  of  the 
circle,  the  angle  now  beijig  formed  by  two  secants. 

Rotating  the  sides  of  the  angle  about  0,  Fig.  207,  until 
they  became  tangent  to  the  circle.  Figs.  208  and  209  are 
obtained. 


176  SECOND-YEAR  MATHEMATICS 

Some  of  the  following  theorems  show  how,  in  each  of 
the  Figs.  203  to  209,  the  measure  of  the  angle  formed  by 
the  two  intersecting  lines  may  be  expressed  in  terms  of  the 
intercepted  arc  or  arcs. 

297.  Measure  of  a  central  angle.  Let  ZAOB,  Fig.  210, 
be  a  central  angle  and  let  it  be  divided 
into  equal  parts.  Taking  one  of 
these  as  a  unit,  the  number  of  equal 
parts  is  the  measure  of  the  angle. 
What  is  the  measure  of  ZAOBf 

Show  that  CD  is  divided  into 
equal  parts. 

Taking  as  a  unit  one  of  the  equal  ^^^ 

' — ^  .  ''"^ 

parts  of  CD,  what  is  the  measure  of  CD  ? 

In  general,  if  the  measure  of  a  central  angle  is  m,  the 
measure  of  the  intercepted  arc  is  also  m.    Why  ? 

Briefly,  we  may  say  a  central  angle  has  the  same 
measure  as  the  intercepted  arc,  or — 

A  central  angle  is  measured  by  the  intercepted  arc. 

EXERCISES 

1.  Draw  a  central  angle.  With  a  protractor  find  the  num- 
ber of  degrees,  integral  or  fractional,  contained  in  the  angle. 
How  many  arc-degrees  are  there  in  the  intercepted  arc  ?  What 
is  a  measure  of  the  intercepted  arc  ? 

.  2.  Draw  a  circle  and  mark  off  an  arc.     Find  the  number  of 
arc-degrees  contained  in  it.    What  is  the  measure  of  the  arc  ? 

3.  Using  ruler  and  compass  only,  divide  a  circle  into  2 
equal  arcs,  4  equal  arcs,  8  equal  arcs. 

4.  Using  ruler  and  compass  only,  construct  arcs  of  90°,  45°, 
60°,  30°,  15°,  75°,  105°,  165°. 

How  may  an  angle  of  90°  be  trisected  ?     An  angle  of  45°  ? 


MEASUREMENT  OF  ANGLES  BY  ARCS 


177 


6.  Divide  a  circle  into  three  arcs  in  the  ratio  1:2:3. 
Find  algebraically  the  number  of  degrees  in  each.    Then  use  the 
protractor  to  draw  the  arcs. 

6.  A  circle  is  divided  into  4  arcs  in  the  ratio  1:4:6:7.  Find 
the  number  of  degrees  contained  in  each  arc. 

7.  The  length  of  a  circle  is  63  inches.  A  central  angle  inter- 
cepts an  arc  7  in.  long.  How  many  degrees  does  the  angle 
contain  ? 

8.  In  the  same  or  equal  circles  two  central  angles  have  the  same 
ratio  as  the  arcs  intercepted  by  their  sides. 

To  show  this,  let  the  measures  of  the  angles  be  m  and  n, 
respectively. 

Show  that  the  measures  of  the  intercepted  arcs  are  also  m  and 
n  respectively. 


Then  each  ratio  is 


Why? 


9.  In  Fig.  211,  AB  is  a 
diameter.  The  member  of  de- 
grees in  ZAOC  is  denoted  by 
x'^-\-4x  and  in  Z  BOC  by  Sx^+  12a;. 
Find  the  values  of  x  and  the 
number  of  arc-degrees  in  arcs  AC 
and  CB. 


Fig.  211 


Fig.  212 


JlO.  In  Fig.  212  ZA5C  is  a  right  angle.  ZABD  =  {2x^-3)°, 
and  Z  DBC=  (lOa:^— 15)°.  Find  the  values  of  x  and  the  number 
of  degrees  in  arcs  AD  and  DC. 

298.  Measure  of  an  inscribed  angle. 
Draw  an  inscribed  angle,  as  ABC,  Fig.  213. 
With  a  protractor  measure  angle  ABC. 
Find   the   number   of  arc-degrees   in  AC- 

How  does  the  measure  of  the  inscribed 
angle  compare  with  the  measure  of  the  arc  ? 

The  following  theorem  shows  how  to  find  the  measure 
of  an  inscribed  angle  in  terms  of  the  intercepted  arc : 


178 


SECOND-YEAR  MATHEMATICS 


Theorem:  An  inscribed  angle  is  measured  by  one-half 
the  arc  intercepted  by  its  sides. 

Let  ABC,  Fig.  213,  be  an  inscribed  angle  inter- 
cepting AC. 

To   prove    that  ABC  is  measured   by 

§ic. 

In  proving  the  theorem  three  cases  are 
considered : 

Case  I.  The  center  of  the  circle  lies  on 
one  side  of  the  angle,  Fig.  214. 

Proof:  Draw  the  radius  CD. 

Denote  the  measures  of  ABC,  ADC,  and 
AC  by  X,  y,  and  x',  respectively,  and  show 
that  ZBCD  =  x. 

Hence,  we  have  the  relation — 

x-\-x  =  y  Why? 

Solving  for  X,       x  =  ^y 

But,  y  =  x'         Why? 


X  —  nX 


Why? 


Case  II.     The  center  of  the  circle  lies  within  the  angle, 
Fig.  215. 

Proof:  Draw  the  diameter  BD. 

x  =  y+z     Why? 


or 


2/  =  |'         Case  I    • 

(/l\) 

z  =  ~          Case  I 
A 

^J^  Why? 

Fig.  215 

x^\x'       Why? 

\ 

MEASUREMENT  OF  ANGLES  BY  ARCS 


179 


Case  III.     The  center  of  the  circle  lies  outside  of  the 
angle,  Fig.  216. 

Proof:   Draw  the  diameter  BD. 


But 


z  =  x+y 
x  =  z-ij 

z' 
'=2 


and 


y' 


2 
1^/ 


Why? 
Why? 

Why? 
Why? 
Why? 


or  x  =  f  X 

299.  Segment  of  a  circle.  The  portion 
of  a  plane  included  between  a  chord  and 
the  arc  it  subtends  is  a  segment  of  the  circle. 
The  shaded  part  ABC,  Fig.  217,  is  a  seg- 
ment of  circle  0. 

EXERCISES 

Prove  the  following  exercises: 

1.  All  angles  inscribed  in  the  same  segment  of  a  circle  are 
equal. 

2.  All  angles  inscribed  in  a  semicircle  are  right  angles.* 

3.  All  angles  inscribed  in  a  segment  smaller  than  a  semi- 
circle are  greater  than  a  right  angle. 

4.  All  angles  inscribed  in  a  segment 
greater  than  a  semicircle  are  less  than  a 
right  angle. 

5.  Two  chords  AS  and  CD,  Fig.  ,218, 
intersect  within  a  circle.  Show  that  i^AEC 
and  BED  are  mutually  equiangular  and 
therefore  similar. 

*  Perhaps  known  and  used  by  Thales;    first  proved  by  the 
Pythagoreans. 


180 


SECOND-YEAR  MATHEMATICS 


Fig.  219 


t6.  {Mathematical  puzzle).  Find  the  error  in  the  proof  of 
the  following  theorem:  From  a  point  not  on  a  given  line  two 
perpendiculars  may  be  drawn  to  the  line. 

In  the  two  intersecting  circles  O  and 
O',  Fig.  219,  diameters  AB  and  AC  are 
drawn  from  A,  one  of  the  points  of  inter- 
section of  the  circles. 

Draw  CB  intersecting  the  circles  in 
points  D  and  E. 

Draw  AE  and  AD. 

ZAEC  is  a  right  angle,  being  inscribed  in  a  semicircle. 

.-.    AEA.CB 
Similarly,  A  ABB  is  a  right  angle. 
.-.    ADLCB 

7.  An  inscribed  triangle  is  a  triangle  whose  vertices  lie  on  a 
circle.    Two    angles    of    an    inscribed 

triangle  are  82°  and  76°.  How  many 
degrees  are  there  in  each  of  the  three 
arcs  subtended  by  the  sides? 

8.  Two  circles  intersect  at  points  A 
and  B,  Fig.  220.  AC  and  AD  are 
diameters.  Prove  that  C,  B,  and  D  He 
in  the  same  straight  hne. 


Fig.  220 


300.  Theorem:  An  angle  formed  by  a  tangent  and  a 
chord  passing  through  the  point  of  contact  is  measured  by 
one-half  of  the  intercepted  arc. 

Let  CDy  Fig.  221,  be  tangent  to 
circle  0,  and  let  AB  he  a  chord  of 
the  circle,  drawn  from  the  point  of 
contact. 

To  prove  that  A  ABC  is  meas- 
ured  by  one-half  of  ABy 


MEASUREMENT  OF  ANGLES  BY  ARCS 


181 


Proof:   Draw  the  diameter  BE. 

Denoting  the  measures  of  A  ABC,  EBA,  and  EBC 
by  Xf  y,  and  z,  respectively,  and  the  measures  of  arcs  BA, 
AE,  and  BAE  by  x\  y',  and  z',  we  have  the  following 

relations: 

z  =  x-{-y.  Why? 

.*.     x  =  z—y.  Why? 


But, 


and, 


y= 

-y. 

Why? 

z 

=  |z',  since  z  = 

=  90° 

and  z'  ■■ 

=  180°. 

x=hz'-b'- 

-w 

-yr 

Why? 

.-.    x= 

=  ix'. 

Why? 

EXERCISES 

1.  A  triangle  ABC,  Fig.  222,  is  inscribed  in  a  circle  and 
ZA  =  57°,  Z5  =  66°.    Tangents  are 
drawn  2X  A,  B,  and  C  forming  the 
circumscribed  triangle  A'B'C.    Find 
the  angles  A',  B',  and  C 


2.  Two  angles  of  a  circumscribed 
triangle  A'B'C  are  70°  and  80°, 
Fig.  222.  Find  the  number  of  de- 
grees in  each  of  the  three  angles  of 
the  inscribed  triangle  ABC. 

3.  The  vertices  of  an  inscribed 
quadrilateral  divide  the  circle  into 
arcs  in  the  ratio  3:4:5:6.  Find  the 
angles  of  the  quadrilateral. 

4.  From  the  point  of  tangency, 
A,  Fig.  223,  of  two  circles  tangent 
internally  two  chords  are  drawn 
meeting  the  circles  in  B,  C,  D,  and 
E.    Prove  jBC  II  DE. 

Draw  the  common  tangent, 


Fig.  222 


Fig.  223 


182 


SECOND-YEAR  MATHEMATICS 


5.  Prove  that  the  tangents  drawn 
from  a  point  to  a  circle  are  equal, 
Fig.  224. 

Problems  of  Construction 

301.  Make  the  following  con- 
structions : 

1.  Upon  a  given  line-segnient  as  a  chord  construct  a  segment  of  a 
circle  in  which  the  inscribed  angles  are  equal  to  a  given  angle. 
Given  the  hne-segment  a  and  an  angle  equal  to  x,  Fig.  225. 


Fig.  224 


Fig.  225 

To  construct  upon  o  as  a  chord  a  segment  of  a  circle  in  which  an 
angle  equal  to  x  may  be  inscribed. 

Construction:  DrawA5  =  a. 

At  ^,  on  AB,  construct  Z.CAB  =  x. 

Dt&w  AE±DC. 

Draw  FE,  the  perpendicular  bisector  of  AB.  It  will  meet  AE 
as  at  E.    Why  ? 

With  E  as  center  and  radius  EA  draw  a  circle.  This  circle  must 
pass  through  B.     Why  ? 

AKB  is  the  required  segment. 

Proof:   Let  Z.ALB  be  any  angle  inscribed  in  segment  AKB. 

Then  AALB  =  iAB.  Why? 

ABAC  =  lAB.  Why? 

.-.     AALB=ABAC.  Why? 

.*.     Z.ALB  is  equal  to  x.       Why? 

Test  the  accuracy  of  the  construction  with  the  protractor. 


MEASUREMENT  OF  ANGLES  BY  ARCS 


183 


2.  Make  the  construction  of  problem  1,  using  a  given  obtuse 
angle. 

3.  On  a  given  line-segment,  construct  a  segment  of  a  circle 
containing  an  inscribed  angle  of  60°*  of  30°;  of  120°;  45°; 
135°;  using  ruler  and  compass  only. 


4.  From  a  point  outside  of  a  circle 
to  construct  a  tangent  to  the  circle. 

Let  A  be  the  center  of  the  given 
circle  and  B  the  given  point  outside  of 
the  cu-cle,  Fig.  226. 

To  construct  a  tangent  to  circle  A 
from  B. 


Fig.  226 


Construction:  Find  the  midpoint  of  AB. 

Draw  a  circle  having  A  5  as  diameter,  cutting  circle  A  at  Z) 
and  E. 

Draw  BD  and  BE. 

BD  and  BE  are  the  required  tangents. 

Proof:  Draw  AD  and  show  that  ZAD5  is  a  right  angle.  Then 
BD  is  tangent  to  circle  A.     Why  ? 

J5.  Euclid's     method     of    solving  / 

problem    4,   as    given    in    Book    III,        / 
Theorem   17  of  his  Elements,  was  as       ; 
shown  in  Fig.  227.  \ 

The    given    circle   is   0   and    the         ^^^^ 
given  point,  A. 

A  concentric  circle  is  drawn  through  Fig.  227 

A.     0    and    A    are    joined    with   OA. 

Where  OA  cuts  the  given  circle,  at  B,  erect  CC  perpendicular  to 
OA.  Connect  C  and  C  with  O.  Join  the  crossing  points,  T  and 
7",  with  A.    AT  and  AT'  are  the  required  tangents.    Prove. 

Tropfke  says  the  mode  of  construction  of  problem  4  first 
occurred  in  1583  in  Thomas  Finck's  well-known  and  valuable  geo- 
metrical work,  entitled  Geometriae  rotundi.  Some  elementary 
geometries  of  the  eighteenth  century  followed  Finck's  construction, 
and  some  followed  EucUd's.    Which  do  you  prefer  and  why  ? 


184 


SECOND-YEAR  MATHEMATICS 


6.  To  draw  a  common  tangent  to  two  circles  exterior  to  each 
other. 

The  number  of  common  tangents  to  two  circles  depends  upon 
the  position  of  the  circles.  If  one  circle  is  entirely  outside  of  the 
other,  Fig.  228,  there  are  four  common  tangents,  i.e.,  two  external 
tangents,  AB  and  CD,  and  two  internal  tangents,  EF  and  GH. 


Fig.  229 


Fig.  230 


@(J? 


Fig.  231    Fig. 232 


If  the  circles  are  tangent  to  each  other  externally,  there  are  two 
external  and  one  internal  tangent,  Fig.  229. 

If  the  circles  intersect,  two  external  tangents  can  be  drawn, 
Fig.  230. 

If  the  circles  are  tangent  internally,  there  exists  only  one  external 
tangent,  Fig.  231. 

No  common  tangent  exists  if  one  circle  lies  entirely  within  the 
other.  Fig.  232. 

Notice  that  in  every  case  the  line  passing  through  the  centers 
of  the  circles  is  an  axis  of  symmetry  of  the  figure. 

Let  A  and  A',  Fig.  233,  be  the  center  of  two  circles  exterior  to 
each  other. 


I.  It  is  required  to  draw 
the  common  internal  tangents. 

Construction:  Draw  AA'. 

Divide  A  A'  into  segments 
having  the  same  ratio  as  the 
radii  (§  176),  and  let  B  be  the 
point  of  division. 

From  B  construct  BC 
tangent  to  circle  A  (problem  4). 


Fig.  233 


BC  is  one  of  the  required  internal  tangents. 


MEASUREMENT  OF  ANGLES  BY  ARCS 


185 


Why? 
Why? 

Why? 


Proof:   Draw  AC.     Draw  A'C'±CB. 

If  it  can  be  proved  that  A'C  is  equal  to  the  radius  of  circle  A', 
then  BC  is  tangent  to  circle  A'  (§74). 

Denoting  the  radii  of  circles  A  and  A'  by  R  and  R', 

im  —  ^n  by  construction. 

Prove      AABCc^AA'BC 
AB  ^AC  ^   R 
■'    BA'    A'C    A'C 

:?.  =  _?_ 

••    R'     A'C 

Prove  that  A'C  =  -B'. 
.*.    BC  is  tangent  to  circle  A'. 
Show  how  to  construct  the  other  common  internal  tangent. 

II.  To  draw  the  external  tangents. 

Construction:  Draw 
AA',  Fig.  234. 

Divide  A  A'  externally 
in  the  ratio  of  the  radii  at 
the  point  B  (§  176). 

Draw  BC  tangent  to 
circle  A. 

BC  is  one  of  the  re- 
quired external  tangents. 

Show  how  to  construct  the  other  external  tangent. 

The  proof  is  the  same  as  for  Case  I. 

In  §§  302,  303  we  find  two  illustrations  of  external  and  internal 
tangents  common  to  two  circles. 

302.  Circular  motion.  Circular  motion  may  be  trans- 
mitted by  means  of  a  belt  running  over  two  pulleys, 
Figs.  235,  236. 

Two  pulleys  whose  radii, 
R  and  r,  are  12  in.  and  5  in., 
respectively,  are  fastened  to 
parallel  shaftings  and  are 
connected  by  a  belt,  Fig. 
235.  Fig.  235 


Fig.  234 


186 


SECOND-YEAR  MATHEMATICS 


The    distance,    a,    between    the    centers    of    the    pulleys    is 
32  inches.    Make  a  drawing  to  the  scale   1  to   16. 

Find  the  length,  I,  of  the  belt 
from  the  formula 

Z=7r(i2+r)+2a. 

In  Fig.  236  the  puUeys  are  con- 
nected by  a  crossed  belt.  Find  the 
length  of  the  belt  by  means  of  the  Fiq  236 

formula. 

l  =  2V  (2^+r)2+a2+7r(i2+r). 

Notice  that  the  pulleys,  as  connected  in  Fig.  236,  turn  in  opposite 
directions. 

303.  Lunar  eclipse.  A  lunar  eclipse  occurs  when  the 
moon  passes  through  the  earth's  shadow.  If  the  moon  is 
within  the  dark  part  of  the  shadow,  Fig.  237,  the  eclipse 


Fig.  237 


is  said  to  be  total.  This  part  is  included  between  the  earth 
and  the  two  external  tangents  common  to  the  earth  and 
the  sun.  If  the  moon  is  in  the  half-light  region  which  is 
determined  by  the  common  internal  tangents  the  eclipse 
is  said  to  be  partial. 

Find  the  length  of  the  earth's  shadow,  taking  the  distance  from 
Earth  to  Sun  as  93,000,000  mi.,  the  diameter  of  the  Sun  as  866,500 
mi.,  and  the  diameter  of  Earth  as  8,000  miles. 


MEASUREMENT  OF  ANGLES  BY  ARCS 


187 


304.  Theorem:   //  two  chords  intersect  within  a  circle, 
either  angle  formed  is  measured  by  one-half 
the  sum  of  the  intercepted  arcs. 

Draw  AD,  Fig.  238. 
Show  that      x  =  y-}-z 

y=W 

z  =  \z' 
.-.     x  =  W-^z') 


305.  Theorem:   //  two  secants  meet  outside  of  a  circle 
the  angle  formed  is  measured'  by 
one-half  the  difference  of  the  inter- 
cepted arcs. 

Draw  AD,  Fig.  239. 
Show  that      y  =  x-\-z 
and  x  =  y—z 

Complete  the  proof  as  in  §  304.  Fig.  239 


306.  Theorem:  The  angle  formed  by  a  tangent  and  a 
secant  meeting  outside  of  a  circle  is  measured  by  one-half  the 
difference  of  the  intercepted  arcs. 

Draw  CD,  Fig.  240.  ^-5^r-r7  ^^b 

Then  y  =  x+z 

and  x  =  y—z 

y=W 


x=W-^') 


Fig.  240 


307.  Theorem:     The   angle  formed   by   two   tangents 
to   a  circle  is  equal  to  one-half 
the   difference   of  the   intercepted 
arcs. 


Show  that 


y=x+z.  Fig.  241. 
x  =  y-z 


=  J(2/'-^0 


188 


SECOND-YEAR  MATHEMATICS 


EXERCISES 

1.  The  arcs  and  angle  being  denoted  as  in  Fig.  242,  find  x 
and  y. 

2.  Find  x  and  y,  Fig.  243,  the  arcs  and 
angle  between  the  secants  being  as  indi- 
cated in  the  figure. 

3.  When  two  tangents  to  a  circle  make 
an  angle  of  60°  into  what  arcs  do  they 
divide  the  circle  ? 


|4.  Into  what  arcs  do  two  tangents  at 
right  angles  to  each  other  divide  the  circle  ? 

5.  Two  tangents  include  two  arcs  of  a 
circle,  one  of  which  is  four  times  the  other. 
How  many  degrees  in  the  angle  they  form? 


Fig.  243 


J6.  The  angle  between  two  secants 
intersecting  outside  of  a  circle  is  76°. 
One  of  the  intercepted  arcs  is  243°.    Find  the  other. 

7.  The  points  of  tangency  of  a  circumscribed  quadrilateral 
divide  the  circle  into  arcs  in  the  ratio  of  7:8:9:12.  Find  the 
angles  of  the  quadrilateral. 

8.  Two  tangents  to  a  circle  from  an  outside  point  form  an 
angle  of  70°.  What  part  of  the  circle  is  the  larger  arc  included 
by  the  points  of  tangency  ? 

9.  The  angle  between  two  secants  is 

30°,  Fig.  244.    The  nimiber  of  degrees  in 

nzP-                 ,  Au    6x^+29x4-30    . 
arc  DE  is  represented  by o^^'i >  i^ 


2x^-7x-15 
x-5 


2;t;+3 
Find  X  and 


the  arc  BC,  by 
the  number  of  degrees  in  each  of  the  two  arcs. 
Reduce  the  fractions  to  lowest  terms. 


Fig.  244 


MEASUREMENT  OF  ANGLES  BY  ARCS 


189 


JlO.  In  Fig.  245  Z.AED  is  60°,  arc  BC  is  represented  by 

a:^+8a;+15  .  ^  ,      x^+12x-^5      _..    ,  . 

^+3       >^^^-^Aby       ^^^3      .    Fmd  a 

the  number  of  degrees  in  each  of  the  two  arcs. 

11.  Prove  that  the  sum  of  the  three  angles 
of  a  triangle  is  two  right  angles. 

In  Fig.  246,  let  ABC  be  any  triangle. 
Circumscribe  a  circle  about  it. 

The  three  inscribed  angles  are  measured  by 
one-half  the  sum  of  the  three  arcs  AB,  BC,  and 
CA. 

But  the  sum  of  the  three  arcs  AB,  BC,  and 
CA  is  the  entire  circle. 

.'.One-half  the  circle,  or  180°,  is  the  measure 
of  the  sum  of  the  three  angles  of  the  triangle. 

1 12.  In  laying  a  switch  on  a  railway  track 
a  "frog"  is  used  at  the  intersection  of  two 
rails  to  allow  the  flanges  of  the  wheels  moving  on  one  rail  to 
cross  the  other  rail.     Show  that  the 
angle  of  the  frog,  a,  Fig.  247,  made 
by  the  tangent  to  the  curve  and  the 
straight  rail  DE,  is  equal  to  the 
central  angle  FOB,  of  the  arc  BF. 


Fig.  246 


MISCELLANEOUS   EXERCISES 

308.  A  limited  number  of  the 
exercises  below  may  be  worked: 

1.  Prove   that   the  circles  de-  Fig.  247 
scribed    on    any    two    sides    of  a 

triangle  as  diameters  intersect  on  the  third  side. 

2.  A  circle  described  on  one  of  the  two  equal  sides  of  an 
isosceles  triangle  as  a  diameter,  cuts  the  base  at  its  middle  point. 

3.  Prove  that  if  a  circle  is  circumscribed  about  an  isosceles 
triangle,  the  tangents  drawn  through  the  vertices  form  an 
isosceles  triangle. 


190  SECOND-YEAR  MATHEMATICS 

4.  A  point  moves  so  that  the  angle  made  by  the  two  lines 
that  connect  it  with  two  fixed  points,  C  and  D,  is  always  the  same. 
Find  the  locus  of  the  point. 

5.  Prove  that  a  parallelogram  inscribed  in  a  circle  is  a  rec- 
tangle. 

6.  Two  lines,  Fig.  248,  are  drawn  /i~^^^^       \'/^ — ^ 

through  the  point  of  tangency  of  two  [  j       ^^"^>1^^'^M 

circles  touching  each  other  externally.  I  }    ^^^"""'^^I^K^'^'^a) 

If  the  lines  meet  the  circles  in  points  b'^^        y\           ^ 

A,  B,  C,  and  D,  prove  AB  li  CD.  ^"""T^  o.o 

Fig.  248 

7.  Two  circles  intersect  at  points  A 

and  B.  A  variable  secant  through  A  cuts  the  circles  in  C  and 
D.  Prove  that  the  angle  CBD  is  constant  for  all  positions  of 
the  secant. 

8.  Two  circles  are  tangent  to  each  other  externally,  and  a  hne 
is  drawn  through  the  point  of  contact  terminating  in  the  circles. 
Prove  that  the  radii  to  the  extremities  of  the  line  are  parallel. 

9.  Given  two  diagonals  of  a  regular  inscribed  pentagon 
intersecting  within  it.  Find  the  number  of  degrees  in  the  angle 
between  them. 

10.  In  triangle  ABC  the  altitudes  BD  and  AE  are  drawn. 
Prove  ZABD=  ZAED. 

Draw  a  semicircle  on  AB  as  diameter. 

11.  One  side  of  a  triangle  is  fixed  in  length  and  position,  and 
the  opposite  angle  is  given.  The  other  two  sides  being  variable, 
find  the  locus  of  the  movable  vertex. 

12.  Two  circles  are  tangent  externally  at  P.  A  tangent 
common  to  the  two  circles  touches  them  at  points  A  and  B. 
Prove  ZAPB=90°. 

13.  Two  circles  are  tangent  externally.  A  line  through  the 
point  of  tangency  intersects  the  circles  at  A  and  B,  respectively. 
Prove  that  the  tangents  at  A  and  B  are  parallel. 


MEASUREMENT  OF  ANGLES  BY  ARCS 


191 


14.  Three  circles,  Fig.  249,  touch  each  other  at  A,  B,  and 
C.  Lines  AB  and  AC  meet  the  third 
circle  at  E  and  D.  Prove  that  E,  0, 
and  Z)  lie  in  the  same  straight  hne. 


Fig.  249 


Fig.  250 


15.  In  Fig.  250  AC  and  DF  are  drawn  through  the  points 
of  intersection  of  two  circles.     Prove  that  AD  \\  CF. 

Prove  a; +2/  =  180,  u +s  =  180 

.*.    x+u  =  lSO 

16.  Prove  that  the  common  external  tangent  AB,  Fig.  251, 
to  two  chcles  that  are  tangent  externally  is  a  mean  proportional 
between  the  diameters  of 
the  curcles. 

Prove  that  AE=EB  =  EF 
is  a  mean  proportional  be- 
tween CF  and  FD. 


Fig.  251 


Fig.  252 


17.  Triangle  DEF,  Fig.  252,  is  formed  by  joining  the  feet  of 
the  altitudes  of  A  ABC.  Prove  that  the  altitudes  bisect  the 
angles  of  ADEF, 

Show  that  x  =  y,  both  being  complements  of  Z.ACB. 

Draw  circles  on  AO  and  BO  as  diameters. 

Show  that  x'=x  and  y'  =  y. 

.'.    x'  =  y\ 


192  SECOND-YEAR  MATHEMATICS 

18.  Prove  that  a  line  from  the  center  of  a  circle  to  the  point 
of  intersection  of  two  tangents  bisects  the  angle  between  the 
tangents. 

Summary 

309.  The  chapter  has  taught  the  meaning  of  the  fol- 
lowing terms: 

inscribed  angle  inscribed  and  circumscribed 

segment  of  a  circle  polygons 

310.  The  following  theorems  were  shown  to  be  true: 

1.  A  central  angle  is  measured  by  the  intercepted  arc. 

2.  In  the  same  or  equal  circles  two  central  angles  have 
the  same  ratio  as  the  intercepted  arcs. 

311.  The  following  theorems  were  proved: 

1.  An  inscribed  angle  is  measured  by  one-half  the  arc 
intercepted  by  the  sides. 

2.  An  angle  formed  by  a  tangent  and  a  chord  passing 
through  the  point  of  contact  is  measured  by  one-half  of  the 
intercepted  arc. 

3.  If  two  chords  intersect  within  a  circle  either  angle 
formed  is  measured  by  one-half  the  sum  of  the  intercepted 
arcs. 

4.  //  two  secants  meet  outside  of  a  circle  the  angle  formed 
is  measured  by  one-half  the  difference  of  the  intercepted  arcs. 

5.  The  angle  formed  by  a  tangent  and  a  secant  meeting 
outside  of  a  circle  is  measured  by  one-half  the  difference  of  th3 
intercepted  arcs. 

6.  The  angle  formed  by  two  tangents  to  a  circle  is  equal 
to  one-half  the  difference  of  the  intercepted  arcs. 


MEASUREMENT  OF  ANGLES  BY  ARCS  193 

312.  The  following  constructions  were  taught: 

1.  Upon  a  given  line-segment  as  a  chord  construct  a 
segment  of  a  circle  in  which  the  inscribed  angles  are  equal 
to  a  given  angle. 

2.  From  a  point  outside  of  a  circle  to  construct  a  tangent 
to  the  circle. 

3.  To  draw  the  common  external  and  internal  tangents 
to  two  circles  exterior  to  each  other. 


CHAPTER  XII 


PROPORTIONAL  LINE-SEGMENTS  IN  CIRCLES 

313.  A  railroad  surveyor  wishes  to  determine  the 
radius  of  a  circular  railway  curve  ABC,  Fig.  253.  He 
measures  the  chord  AC,  and  BD,  the 
part  of  the  perpendicular  bisector  of  AC 
intercepted  by  AC  and  arc  ABC,  If 
AC  =  200  ft.  and  BD  =  Q  ft.,  how  may  the 
radius  be  determined  ? 

If  we  can  estabUsh  a  relation  between 
AD,  DE,  DC,  and  DB,  the  problem  will 
easily  be  solved. 

To  find  this  relation,  draw  a  circle. 
Fig.  254,  and  a  chord  AC  intersecting 
chord  BE,  as  at  D.  Measure  to  two 
decimal  places  the  segments  AD,  DE,  DC, 
and  DB  and  compare  AD -DC  with 
ED'DB. 

Note  the  approximate  equality  of  the 
products  of  the  segments  of  each  of  the  Fig.  254 

two  chords. 

To  what  is  the  difference,  if  any,  probably  due  ?    This 
illustrates  the  following  theorem : 


E 

Fig.  253 


314.  Theorem:  //  two  chords  of  a  circle  intersect,  the 
product  of  the  segments  of  one  is  equal  to  the  product  of  the 
segments  of  the  other. 

State  the  hypothesis  and  the  conclusion.  Then  prove 
the  theorem  as  follows: 

194 


PROPORTIONAL  LINE-SEGMENTS  IN  CIRCLES     195 


Proof:   Draw  BC  and  AE,  Fig.  255. 

Prove  AADEc^ABDC. 

^,        ^,    .  AD    DE 
bnow  tnat  7775  =  7;^. 
JJn     UL> 

,\    AD'DC=DE'DB.        Why? 


EXERCISES 


Fig.  255 


1.  Solve  the  problem  of  §  313  by  applying  the  theorem  in 
§314. 

2.  Using  the  theorem  in  §  314,  construct  a  square  equal  to  a 
given  rectangle. 

In  a  circle  large  enough  draw  a  chord  equal  to  the  sum  of  two 
consecutive  sides  of  the  given  rectangle. 

Draw  a  radius  to  the  point  of  division.  What  chord  through 
this  point  is  bisected  at  the  point  ? 

3.  Show  how  exercise  2  may  be  used  to  find  geometrically 
the  square  root  of  a  number.  Using  this  method,  find  the  square 
roots  of  6;  5;  10. 

4.  The  segments  of  two  intersecting  chords  are  x-{-5  and 
a; -^6  of  the  one,  and  x-\-2  and  a;— 5  of  the 
other.    Find  x  and  the  length  of  each  chord. 

5.  A  chord  of  a  circle  DC,  Fig.  256, 
cuts  the  chord  AB  at  the  midpoint  E.  ED 
is  4  in.  longer  than  EC  and  AB=1Q  inches. 
Find  the  lengths  of  ED  and  EC  approxi- 
mately to  TOO  inch. 

6.  The  segments  of  intersecting  chords 
are    given    below.    Find   x. 


Fig.  256 


First  Chord 

Second  Chord 

1 

X-4: 

a;+8 

a;+3 

x-4: 

2 

x+2 

a;+6 

a;-4 

a;+18 

t3 

2a; +5 

a;+l 

a;+2 

3a; +2 

14 

2a; +2 

3a; -5 

x-\-l 

a;+5 

196 


SECOND-YEAR  MATHEMATICS 


|7.  The  distance  between  two  points,  A  and  B,  on  a  railroad 
curve  is  2a  ft.,  and  the  distance  from  the  midpoint  of  the  chord 
AB  to  the  midpoint  of  the  curve  is  h  feet.    Find  the  radius. 

|8.  Find  the  radius  of  the  circle  in  exercise  7  if  a  =  100,  6  =  4; 
a=150,  6  =  5.6. 

9.  How  far  in  one  direction  can  a  man  see 
from  the  top  of  a  mountain  2  mi.  above  sea- 
level? 

Let  AB,  Fig,  257,  represent  the  height  of  the 
mountain  and  let  AD  be  the  required  distance. 

Assuming  the  diameter  of  the  earth  to  be 
8,000  mi.,  the  value  of  AD  may  be  found  if  we 
establish  a  relation  between  AB,  AD,  and  AC. 


Fig.  257 


The  following  theorem  expresses  this  relation: 

315.  Theorem:  If  from  a  point  without  {outside  of) 
a  circle  a  tangent  and  secant  he  drawn,  the  tangent  is  a  mean 
proportional  between  the  entire  secant  {to  the  concave  arc) 
and  its  external  segment. 

State   the  hypothesis   and   the 
conclusion. 

Proof:  Draw   DB   and   DC, 
Fig.  258. 

Show    ^ABDc^AACD. 
AC^^AD 
"    AD    AB' 


Why? 


EXERCISES 

1.  Using  the  theorem  §  315,  solve  exercise  9,  §  314. 

2.  If  two  adjacent  sides  of  a  rectangle  are  given,  show  how 
the  theorem  in  §  315  may  be  used  to  construct  other  equivalent 
rectangles. 

3.  Using  the  theorem  in  §  315,  show  how  to  construct  a  square 
equal  to  a  given  rectangle. 


PROPORTIONAL  LINE-SEGMENTS  IN  CIRCLES     197 


4.  Show  how  the  theorem  in  §  315  may  be  used  to  find 
geometrically  the  square  root  of  a  number. 

5.  Prove  by  means  of  the  theorem  in  §  315  that  the  two  tan- 
gents from  an  external  point  to  a  circle  are  equal. 

6.  A  tangent  and  a  secant  are  drawn  from  the  same  point 
outside  of  a  circle.    The  secant 

measured  to  the  concave  arc 

is  three  times  as  long  as  the 

tangent,  and  the  length  of  its 

external  segment  is   10  feet. 

Find  the  length  of  the  tangent 

and  secant.  „ 

Fig.  259 

7.  Using  Fig.  259,  prove 

that  the  square  of  the  hypotemise  of  a  right  triangle  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides. 

Let  ABC  be  a  right  triangle  having  ZC  =  90°. 

Show  that  BE'BD=BC^. 

Hence,  (c+6)(c-6)  =a\  or,  c'^a^+b^ 

8.  To  divide  a  line-segment  into  two 
parts  so  that  the  longer  part  is  a  mean 
proportional  between  the  whole  segment 
and  the  shorter  part. 

Let  AB  be  the  given  line-segment, 
Fig.  260. 

To    find    the    point    C,    such    that 
AB^AC 
AC     CB' 

Construction:  Draw.BD±A.S  at  B,  making  BD='—  • 

With  D  as  center  and  radius  Z)B,  draw  circle  D. 
Draw  AD  cutting  circle  D  at  E  and  F. 
On  AB  \a.yofl  AC  =  AE. 
C  is  the  required  point. 

This  may  be  proved  as  follows : 


198  SECOND-YEAR  MATHEMATICS 


>,      .    AF    AB  „,^    ^ 


AF-AB^AB-AE 

AB  AE 

AF-EF    AB-AC 


(§  195) 
Why? 
Why? 
Why? 
Why? 


AB  AC 

,  AE^CB 

"  AB    AC 

,  AC_CB 

"  AB    AC 

,  AB^AC 

••  AC    CB' 

Problem  7  will  be  used  in  the  construction  of  a  regu- 
lar inscribed  decagon  (10-side),  §  443. 

316.  Mean  and  extreme  ratio.*  A  line-segment  is 
divided  into  mean  and  extreme  ratio  if  the  longer  part 
is  a  mean  proportional  between  the  segment  and  its 
shorter  part. 

*  The  current  method  of  dividing  a  line  in  extreme  and  mean 
ratio  is,  according  to  an  Arabian  commentator,  due  to  Heron  of 
Alexandria.  The  theorem  for  dividing  the  line  has  been  called  by 
various  names.  Plato  called  it  ''The  Section";  Lorentz  (1781) 
called  it  "Continued  Division." 

Campanus  (last  half  of  the  tweKth  century)  called  continued 
division  ''a  wonderful  geometrical  performance."  Paciolo  (1445- 
1514)  gave  it  even  higher  esteem  by  writing  an  entire  work  dealing 
with  problems  in  continued  division  and  gave  his  work  the  title: 
Divine  Proportion. 

The  peculiar  mysticism  of  later  times  seized  upon  Paciolo's 
idea  and  went  still  beyond  him.  Ramus  (1515-72)  associated  the 
divine  trinity  with  the  three  segments  of  a  continued  division. 
Kepler  (1571-1630)  created  a  complete  symbolism  for  his  sectio 
divina  ("divine  section").  In  the  middle  of  the  nineteenth  century 
there  arose  a  sort  of  amateurish  natural  philosophy  that  sought  to 
subtihze  mathematical  laws  in  every  branch  of  study.  A  kind  of 
universal  validity  was  fantastically  ascribed  to  this  continued  divi- 
sion, and  it  was  now  christened  "Golden  Section." 

This  "Golden  Section"  was  held  to  be  not  only  the  criterion 
for  all  metrical  relations  in  nature,  but  it  was  also  regarded  as 
the  "principle  of  beauty"  in  painting,  architecture,  and  the 
plastic  arts,  as  well.  (Tropfke,  Geschichte  der  Elementar-Mathematik, 
Band  H,  S.  99-103.) 


PROPORTIONAL  LINE-SEGMENTS  IN  CIRCLES     199 

317.  Theorem:  //  from  a  point  without  a  circle  two 
secants  are  drawn  to  the  concave  arc,  the  product  of  one 
secant  and  its  external  segment  is  equal  to  the  product  of 
the  other  secant  and  its  external  segment. 


Fig.  261 

Proof:  From  C  draw  CF,  Fig.  261,  tangent  to  the 
circle. 

Show  that  CA  -CB^CF^  and  CD  •  CE=CF\ 

EXERCISES 

1.  Two  secants  to  the  same  circle  from  an  outside  point  are 
cut  by  the  circle  into  chords  that  are  to  their  external  segments 
as  |-  and  5(  =  |^) .  The  first  secant  is  8  ft.  long.  Find  the  length 
of  the  second  secant. 

2.  The  following  exercises  relate  to  two  secants  from  an 
external  point  as  in  exercise  1.  Find  the  length  of  the  second 
secant. 


Ratios  of 
Segments  of 
First  Secant 

Ratios  of 

Segments  of 

Second 

Secant 

Length  of 
First  Secant 

1 

5:2 

3:1 

28  ft. 

2 

3:1 

5:2 

28  ft. 

t3 

4:1 

5:4 

625  ft. 

1:4 

4:1 

4:3 

25  ft. 

|5 

7:2 

7:3 

36  ft. 

200 


SECOND-YEAR  MATHEMATICS 


J3.  Two  lines  drawn  through  the  common  points  of  two 
intersecting  circles,  Fig.  262,  meet  the  circles  in  A,  B,  C,  and 
Z),  E,  F,  respectively.    Prove  AD  \\  CF.  ^ 

GA    GD 


Show  that 


GC    GF 


Fig.  262 


r.-^-G 


Fig.  263 


4.  Show  how  to  find  a  point  such  that  the  tangents  to  two 
given  circles  are  equal  (see  Fig.  263). 

6.  Determine  a  point  A  without  a  circle  so  that  the  sum 
of  the  length  of  the  tangents  from  A  to  the  circle  shall  be  equal 
to  the  distance  from  A  to  the  farthest  point  of  the  circle. 

Summary 

318.  The  following  theorems  were  proved: 

1.  If  two  chords  of  a  circle  intersect,  the  product  of  the 
segments  of  one  is  equal  to  the  product  of  the  segments  of  the 
other. 

2.  If  from  a  point  without  a  circle  a  tangent  and  secant 
he  drawn  the  tangent  is  a  mean  proportional  between  the 
entire  secant  to  the  concave  arc  and  the  external  segment. 

3.  If  from  a  point  without  a  circle  two  secants  he  drawn 
to  the  concave  arc,  the  product  of  one  secant  and  its  external 
segment  is  equal  to  the  product  of  the  other  secant  and  its 
external  segment. 

319.  The  following  construction  was  taught: 
To  divide  a  segment  into  mean  and  extreme  ratio. 


CHAPTER  XIII 

THE  OPERATIONS  WITH  FRACTIONS.    FRACTIONAL 
EQUATIONS 

320.  In  future  work  we  shall  need  considerable  skill 
in  working  with  fractions,  which  occur  in  many  problems. 
It  is  the  purpose  of  this  chapter  to  review  and  extend  our 
knowledge  of  the  operations  with  fractions. 

Addition  and  Subtraction  of  Fractions 

321.  Adding  and  subtracting  fractions  that  have  the 
same  denominator. 

EXERCISES 


1.  Show  from  Fig.  264  that 


3    4^7  H— ^^/^— -. i 


-I % , 

i — I — I 1 — h 


7/  1 

2.  Show  from  a  figure  that      '  ^8  '-^^h 


7_4^3 
9    9    9 


Fig.  264 


3.  Make  a  rule  for  adding  and  subtracting  fractions  having 
the  same  denominator  and,  using  this  rule,  combine  each  of  the 
following  expressions  into  a  single  fraction: 

7"^7  13"^  13     13     13 

^"5^5  ^'9    9    9^9 

o   1^4    2  ^  7    5    3     11 

3     3    3  2     2     2     2 

201 


c    c 

8."-^ 

c     c 

9.  ^+5+^ 
X     X    a; 

10.  r__^+^_.^ 

y   y   y   y 

11     3  +  ^  - 

7 

a+6^a+6 

a+& 

12      ^        ^^^ 

31?/     43x 

•  16m     16m 

16m     16m 

202  SECOND-YEAR  MATHEMATICS 

c—ax,c-\-3ax 

13.     — -; 1 

4s  4s 

3a    3a -86 
56        56 

10a;  ,  12x_  5x 
'  17a'^17a     17a 

,^   5a-6    2a-36 
a;-?/       x-y 

/61a  ,  48c\  _ /49a    57c\ 
Vl76'^176/     Vl76"^176/ 

17i/+18^_22y+llg 
27a;  27a; 

/29a;     132/\      /33a;     192/\ 

^^'  \io'~lo)'^\'io~lo) 

2x-hl5y    9x-Sy_7x-Sy 
QK     "^    6K  QK 

23a+86_19a-286_176-8a 
■      12a;  12a;  12x 

322.  Adding  and  subtracting  fractions  having  different 
denominators. 

EXERCISES 

1.  Reduce  f  and  f  to  fifteenths. 

2.  Reduce  f  and  f  to  fractions  having  the  same  denominator. 

3.  Add  I  and  |. 

5    7^5 ♦43- 7^5  -  4+3 • 7^  82  ^41 , 
6"^8    6  •  4"^3  -8  6-4       ~6  •  4    24  ' 

4.  Subtract  f  from  f . 

8    4^7-8    9 -4^7 > 8-9 -4^20 
977.99-7  9.7  63* 


FRACTIONS.  FRACTIONAL  EQUATIONS    203 

323.  Exercises  1  to  4,  §  322,  show  that  fractions  with 
different  denominators  are  added  (or  subtracted)  by  first 
changing  the  form  so  that  all  have  the  same  denominator. 
The  sum  (or  the  difference)  of  the  numerators  is  then  written 
over  the  common  denominator  and  the  resulting  fraction 
reduced  to  its  lowest  terms. 

EXERCISES 

In  the  following  exercises  change  to  one  fraction  each  of 
the  indicated  sums  and  differences,  giving  as  many  as  you  can 
mentally.  Reduce  all  results  to  lowest  terms  by  dividing 
numerator  and  denominator  by  common  factors. 

*  15    20    9    4"^  18  *  14    4    3'^21 

2.  ?_?4-2?-l^  4.  0:4-—-—+- 

3    8^  4       12  ^^22    33^6 

6.  2|,+4^.-8.+3A.+  l|. 


®*   \4a'^5a)     \6c 


5 2_- 

Qa     18a    5a> 


7    ^-L^  13    <*~26    4a— 56 


8. 


9. 


ex    cy 

1_^ 
c    ca 

X    _  z 
12ab    66 


3a; 

5x 

14. 

1 

16. 

2  _ 

xy 

Sy^^xy'-\-y' 
xy^       xhf 

ifi 

a 

ab 

10.  ^+^4-^ 

a6        be        ca  a—1     a{a—l) 


x^y    xy^  "       x-l     2(x-l) 

t^-U^LZ^  18        1       ■     x+y 

X    ^  Sx  *  2x-Sy~^4x^-Qxv 


204 


SECOND-YEAR  MATHEMATICS 
2        5 


19        ^     4-       ^^ 
*  a+2b^3ad+6bd 


20.  -+& 

X 


Put  6: 


21.  ?-a 


t22.  5a+ 


2a 


23.  ^+^+^ 
DC    ac    ao 


|24. 


1        1 


x-y 


2g^  5a;-4j/+3g  ^  2x+Sy-4:Z 
X  dy 

126    '^^+3^~4^_^^+^^~^^ 


27.  ^-^+-^ 

2-2^        2-2^2        3.^2 


28. 


29. 


3    a-\-b 

a-\-b_    a 
a       a—b 


30.  ^J^+^-l 

n^  X        ^^X~\~0        X  —  O 

x^—l     x  —  l     x+l 


t32. 
33. 


Za  _    a    _   2ac 
c+d    c^    c^-d"^ 

1 1_ 

1  1 


«}4. 

a+&     a 

^,-b 

3R 

1 

2a+36 

1 

3a+26 

36 

1 

1 

X2  +  2/2 

a;2-7/2 

<17 

1 

1 

a+6    a+&+c 


38. 
39. 
40. 
41. 

J42. 


3a: 


52/ 


72 


4a(x4-2/)     3<2(a;+2/)     2aa:+2a?/ 

9  7,2 

2a;+4?/    3x+6i/    Sx+lO?/ 

4  3 


(a+l)(a+2)     (a+2)(a+3)     (a+3)(a+l) 

x+4 x-\-2         ,  X 

(x+2)(a;+3)     (^3)7^+4)"^(x+4)(a;+2) 

x+y-g      ,      x-y+g     _    x-\-y-^z 
(x+2)(?/+2)     (x+?/)(?/+2)     (a;+2/)(x+2) 


FRACTIONS.  FRACTIONAL  EQUATIONS    205 


5x-\-7m  _      25x      ,       7x 
4x-\-12m    Qx+lSm'^2x-\-Qm 

44.  J^  +  _^^ ^ 

X  +  IJ      x2-?/2       IJ-X 

Let  y—x=—(x  —  y) 

+415      Sa'^      2a+l_2a-l 
^       a2-l"^2a-2    2a+2 

2a;+3_x'^-lla;+18     a;-6 
x-6  a;2-36        a;+6 

^       25^2-9    5x-3^5x4-3 

^„    x-\-ma  .  x—ma       w?ah 
48.  z h^  ~ 


49. 


hm—x    bm+x    b^m'^—x'^ 

5x-&y         x-\-lSy        SSx''-2xy+15y^ 
6x2+6x2/     lOxy-lOy^        15x'-  15x?/2 


t50.    aj'-2/' _L^-2/_a^'+2/' 


51. 


|62. 


(x+2/)2    x-\-y    X2-2/2 
3        4x+12  6 


2x-3     4x2-9     4x2+12x+9 

2x  Sy     _2x2+3xy-2i/2 

x-\-2y    4uC+8y     x2+4x?/+4?/2 


63.^+     3      •  2 


3x+2  '  5x-l     (3x+2)(5x-l) 

Multiplication  of  Fractions 

2 
--means   ^. 1 

11  ^  11 


2 
324.  The  number  4- —means   ^ 1 i 


2,2,2,2      8  ''■'''''''"" 

11^11^11^11      11  ^Jj.....4,l H 

Thus,  4  •  ^  =  ^    (See  Fig.  265.)  Fig.  265 


206  SECOND-YEAR  MATHEMATICS 


EXERCISES 

1.  Give  the  meaning  of  c  •  , 

0 

2.  Express  in  words  the  equation  c  -^=^^1^ 

0  0 

3.  Multiply  I  by  8;  by  12;  by  5;  by  25;  by  a;  by  xy 

4.Multiplylbyl;lbyl;|byJ;lbyl;lby|;Hy-; 

(2  hy  i,  J  of  J,  Jx|,  and  J  •  J  are  all  equivalent.) 

6.  Multiply  I  by  ^;  by|  by|;  by|;  by?;  by?;  by? 

6.  State  the  rule  for  multiplying  two  fractions  and  compare 
it  with  the  following: 

Fractions  are  multiplied  by  multiplying  their  numerators  for 
the  numerator  of  the  pr9duct,  and  multiplying  their  denominators 
for  the  denominator  of  the  product. 

Since  the  product  of  fractions  should  generally  be  reduced  to 
the  simplest  form,  factors  that  are  common  to  numerator  and 
denominator  should  be  divided  out  before  multiplying. 

7.  Multiply  If  by  I 

12     15^12-  15^3  •  3 

35  *  16     35  .  16    7  •  4 '  ^^' 

8.  Multiply  ^  by  15y^ 

\jy 

??.15^2  =  Z^^l%!  =  7x^    etc 
9y    ^^^  9y  3      '  ^^^• 

9.  Multiply  2^^  by  (2a:+2) 

7x  7a;(2x+2)^  7x  •  2{x+\)  ^  7x 

2x2-2  ^^^^^^       2x2-2       2(a;+l)(x-l)     x-1 


FRACTIONS.    FRACTIONAL  EQUATIONS  207 

^n    lix  u-  1     56x2        IQy 
10.  Multiply  — by  2^ 


56x^    102/^56x2  ♦  lOy 
55y^  '  2\x    55i/2  •  21a; ' 


etc. 


ii    TVT  u-  1    3x+3t/,      2x2-22/2 
11.  Multiply  ^  by  3^ 

3x+3y    2x^-22/2  ^  (3a;+3y)  (2x^-21/2) 
2X-22/  *  3x2+32/2     (2x-22/)(3x2+32/2) 
_3(x+2/)2(x+2/)(x-y)     g^^ 
2(x-2/)3(x2+2/2)       ' 

325.  Exercises  7  to  11,  §  324,  show  that  fractions 
may  be  multiplied  by  writing  the  indicated  products  of 
the  numerators  over  the  indicated  products  of  the  denomina- 
tors and  then  reducing  the  fraction  obtained. 


EXERCISES 

The  following  products  are  to  be  given  in  simplest  form. 
Special  effort  should  be  made  to  cover  this  hst  of  exercises  in  the 
minimum  amount  of  time. 

Multiply  as  indicated: 

1    2    3    1  ^   7xyz    f,  , 

68^    _95^  ^    ah    yz 

'  102    133  xy    be 

3.  1  .  ac  9.  1^  .  g? 

a  3x2/    002/ 

.    1_      3  15ab  .  24x2/g 

*  a2  •^"'  16x2/     256c 

-   a    1  ^^    Sab    5bc    7xz 

5,  —  •  —  11,  •  — -  •  

6    X  4x2/    Qyz    Sac 

Q   ac  ^2    2a^  ^  66^  ^  56 

*  c  '  d  *  Sb'^y  '  7ox2 '  4a 


208 


SECOND-YEAR  MATHEMATICS 


13. 


14. 


16. 


19. 


20. 


21. 


22. 


a±b^a^-¥ 
a-b  '  02+62 

27a;     ^  x+y 
8y+Sx  '     3 

a    ^     h 
a+6    a—b 

18a;2+12a:?/+2!/2 
Sx^-27xy^ 

9a%x—9a%y  ^    4xyh—4:xy^u 
Scx'^u — Scx'^v     loab'^y  —  1 5ab'^x 

6ma+6m&  ^  las— lbs 
35na-35n6  *  9ar+96r 

27pq'm—27pqn  ^  7bpx—7bpy 
S5abx—35aby     9anq—9amq 


5x^y—15xy^ 
36x2|/-4^3 


16. 


17. 


18. 


Q{x-y)      15x^y^ 


bxy^       8{x-y) 

a^—ab    x'^-\-xy 
x^—xy    a^-i-ab 

x^-xy  ^  {a+by 
da+Zb  *  {x-yy 


23. 


X+4: 


x—4i    a;2+4x+4 


Division  of  Fractions 

326.  To  divide  a  number  by  a  fraction  means  to  find 
the  number  which  multiplied  by  the  divisor  gives  the  dividend. 

4  4 

Thus,  6  4-  -  means  to  find  what  number  multipUed  by  -  will 
y  y 

/      9\     4  9 

give  6.    Since  i^'j)  *  q  gives  6,  it  follows  that  6  •  -  is  the 

A  o 

required  number. 


Therefore  6-h^  =  6 
y 


EXERCISES 

1.  Using  the  same  reasoning  divide  the  following  numbers 
by^:  3;  11;  a. 

2.  Similarly  show  that  ©■^o'^c  *  ^ 

o      o      o      O 

3.  Show  that  ?^^  =  ?-- 

babe 


4.  Translate  the  equation  of  exercise  3  into  words. 


FRACTIONS.    FRACTIONAL  EQUATIONS  209 

327.  Reciprocals.  Two  numbers  whose  product  is  1 
are  reciprocals  of  each  other. 

1.  Give  the  reciprocals  of  4,  3,  4^,  |. 

2.  Compare  your  statement  of  exercise  4  with  the 
following: 

A  number  is  divided  by  a  fraction  by  multiplying  the 
dividend  by  the  inverted  divisor;  that  is,  by  multiplying  the 
dividend  by  the  reciprocal  of  the  divisor, 

EXEKCISES 

1.  Divide  25x2  by  ^ 

y 

25x-^=25x^.^=?^,etc. 
y  15a;      15x 

2.  Divide  3— by  ^ 

62x^  .  93xy^Q2x^    55ap ^Q2x^  -  55ap 
35p2  •  55ap  ~35p2  *  Q3xy    35p2  .  9Sxy ' 

4 

3.  Give  in  the  simplest  form:    -r^" 

2^ 
ab 

_4 ab    2a 

^  -  4  X  ^2 ,  etc. 

2a 

Find  the  results  in  the  following  indicated  divisions  and 
reduce  them  to  the  simplest  forms: 

4   3^2  7^6  a^c 

5*7  *  9  •  7  b'  X 

B   4^6^  „   3^7  c_,d 

7  *  11  8*5  '  x'  f 


210  SECOND-YEAR  MATHEMATICS 


10.  x-^y  16.  ^^^'^'  -  ^'^ 


11  a6^^*  17    20^3_^j4a^ 

*  22  •  21a4cs  '  3a262c2 

12  12a;3^1^  18    i0aW^35aW 

7y  '  22mHh^ '  SSm^xz'' 

13.  25a»^i5f  19.  M!^?!-l 

U.  l&W^i^  20.  ^+5±1 

^     5a6  x-4:    x2-16 

22.  (32xV22_40a;2^322)^8xY£ 

23^  24:{x-iy  .        30(x-l) 


70(a2-62)  •  28(a-6)2(a+6) 

14x2-7x    .  2a:- 1  ^^+^)^  •  fg^    5^^ 

•  12x3+24x2  •  x2+2x  a-6    *  ^       ^ 


3(x2-4j/2)^a;-2i/  3x2-3^  1+x 

'    4(a2_62)    •  a-\-h  '    x+y   '  x'^-y' 


Complex  Fractions 

328.  Complex  fractions.    A  fraction  in  which  either 
numerator  or  denominator,  or  both  terms,  contain  frac- 


tions is  a  complex  fraction,    e.g.,  j,  ~^,  and  j. 

T     2  -g- 

The  last  fraction  may  be  reduced  in  two  ways: 


(1)  f =1X1,  etc.,  or  (2)  I  =  fli:^=Ll|,  etc. 


FRACTIONS.  FRACTIONAL  EQUATIONS    211 

In  the  second  method,  numerator  and  denominator 
of  the  complex  fraction  have  been  multipUed  by  the  same 
number,  viz.,  the  least  common  multiple  {l.c.m.)  of  4  and 
9.  Sometimes  the  use  of  this  method  is  more  advantage- 
ous than  the  first  method. 

EXERCISES 

Reduce  the  following  complex  fractions: 


1. 


x-1 

X 

1 

X  — 
X 


x-\     x-l     ^ 

•  X 

X  X  x—l       , 

— r ,  etc. 


X—-     (x — V 
X     \      xj 


2.3  4a;  ,^  3a:  ,  7y 

6  5  6 

1,1  ?    I 

1^1  '  ? _|_ J     ^_i  '   Z     X     X 

y   X  y        y  2 

Perform  the  indicated  operations:     - 

8.  ("2+£)('!L_P) 

\n     0/  \m     0/ 


q/  \m     q^ 

y^     'Zlx^/  '  \y 


Q    /8x3 tS\^(2x_y^\ 

'   \y^     21  x^)  '  \y     3x/ 

10    /2x+5i/    5x-\-2y\  ,  /2x-Sy    7x-Sy\ 
\3x-\-y      x-\-Sy/  '  \  ,  2a;    '^2x+Qy/ 


212  SECOND-YEAR  MATHEMATICS 

Fractional  Equations 
329.  Solve  the  following  equations: 

Multiplying  each  term  by  the  least  common  multiple  of  the 
denominators,  i.e.,  by  12, 

m     12^^12x 
3-1-4        6  ^ 

Reducing,  we  have  4:X+3x=2x+AS,  which  is  easily  solved. 

2.  -2—4-^3 

3.  5x-^^  =  2x-{-2i 


4.  .05(20a;-3.2)=.8(4a;4-.12)-11.25G 

Clear  of  fractions  by  multiplying  every  term  by  100. 

.    .    .       ^   ^.      .21a;+.012     ,   „ 

5.  1.4a;  — 1.61 =  l.Sx 

.0 

l-2a;_2a;-.5  ,  2a;-|_6.35-.5a; 
*     .25        12.5  5  3 

7.  5r-13  =  ?!:f-^+^ 
4  4 


_    5r^-3r+12_.,^(3r+l)(r-10) 
8.  ^  10 


r  ,  1      _r 
.     3"^2_'"    3_     3 
^'•"5 2 2 


FRACTIONS.    FRACTIONAL  EQUATIONS  213 

12      10^  _   lOa^  ^^  13    x-4:_^x-l 

*  2a;-2    Sx-S  '  x-2    x+3 

l.c.m.  =  2  •  Z{x-l) 

Since  exercises  13-20  are  proportions,  the  theorem  that  the 
product  of  the  means  is  equal  to  the  product  of  the  extremes  may  be 
used  to  clear  them  of  fractions. 

14    3a;-l_3a:+l  10^-2^4^+5 

*  4a;+2    4x+5  *  lOx+6    4a;+i 

ti5   ^+3_a;  — 1  ^-    x-f-5_5a;  — 19 

tie   ^±?=^±?  20   4-a:_15-a; 

■^    *  a;+4    x+4  *  1-a;     3-a; 

17    3;+l.l_a;-1.7  ^       5.7;-3_a;+l 

*  x-1.4     X+.3  *    •  5a;+3    x+3 


330.  Summary  of  the  laws  of  fractions. 

State  the  laws  that  the  foUowing  expressions  formulate: 


^    m.n    m-\-n 
•  d'^d'    d 

f.    m    n    m  —  n 

d     d        d 

m_^n    m^n 

d     d        d 

a_^c    a  '  d=^c  • 

b 

b    d         b'd 

6.5.^  =  1 
y   X 

^'b'b 

7. 

n  ■ 

X    n  '  X 

y     y 

8. 

a 
b' 

X    a  '  X 
y     b-y 

9. 

a  . 

r 

a-x-n 

10. 

b' 

a 

rn=:r 

b  •  n 

11. 

h- 

a          b 

'-T=n  •  - 

b          a 

12. 

,  c    a    d 
''d    b'  c 

214  SECOND-YEAR  MATHEMATICS 

Problems  Leading  to  Fractional  Equations 

331.  Motion  problems.  Solve  the  following  prob- 
lems: 

1.  The  report  of  a  cannon  shot  was  heard  3 . 4  seconds  after 
the  flash.  If  sound  travels  1,080  ft.  per  second,  how  far  away 
was  the  cannon  ? 

The  time  it  takes  light  to  travel  1,080  ft.  is  too  smaU  to  be 
considered  in  the  problem. 

2.  In  one  year  light  travels  a  distance  63,000  times  as  great 
as  the  distance  of  the  earth  from  the  sun.  Assuming  the  distance 
of  the  earth  from  the  Pole-star  to  be  2,898,000  times  as  great  as 
the  distance  of  the  earth  from  the  sun,  how  long  does  it  take  the 
light  of  the  Pole-star  to  reach  the  earth  ? 

3.  Two  trains  go  from  P  to  Q  on  different  routes,  one  of 
which  is  15  mi.  longer  than  the  other.  The  train  on  the  shorter 
route  takes  6  hours,  and  the  train  on  the  longer,  running  10 
mi.  less  per  hour,  takes  8|  hours.  Find  the  length  of  each 
route. 

For  the  train  on  the  short  route:     For  the  train  on  the  longer  route: 


d=x 

/ 

d  =  x+\^ 

<  =  6 

t=^\ 

X 

'•• 

x+\b 

.-. 

X 

6" 

-10  = 

x  +  lo 

4.  A  robber  attempted  to  escape  in  an  automobile  going  at 
the  rate  of  28  mi.  an  hour.  Fifteen  minutes  later  he  was 
followed  by  the  police  in  an  automobile  going  at  the  rate  of 
32  mi.  an  hour.    How  soon  did  they  overtake  the  robber  ? 

6.  The  distance  from  A  to  B  is  100  mi.  A  train  leaving 
A  at  a  certain  rate,  meets  with  an  accident  20  mi.  from  B, 


FRACTIONS.    FRACTIONAL  EQUATIONS  215 

reducing  the  speed  one-half  and  causing  it  to  reach  B  1  hour  late. 
What  was  the  rate  per  hour  before  the  accident  ? 

To  solve  the  problem  find  a  relation  between  the  regular  time, 
the  time  before  the  accident,  and  the  time  after  the  accident. 

6.  A  man  walks  beside  a  railway  at  the  rate  of  4  mi.  per 
hour.  A  train  208  yd.  long,  running  30  mi.  per  hour,  over- 
takes him.    How  long  will  it  take  the  train  to  pass  the  man  ? 

7.  Two  boys  are  running  along  a  circular  path  whose  length 
is  100  feet.  When  they  run  in  opposite  directions,  they  meet 
every  eight  seconds,  and  when  they  run  in  the  same  direction 
they  are  together  every  25  seconds.    What  are  their  rates  ? 

•    332.  Percentage  and  interest  problems. 
Solve  the  following  problems: 

1.  A  property  owner  uses  8  per  cent  of  the  money  received 
for  rent  to  pay  the  taxes.  His  taxes  having  been  raised  to  11  per 
cent,  what  per  cent  must  he  raise  the  rent  in  order  to  keep  his 
income  the  same  as  it  was  before  ? 

Denoting  by  A  the  amount  received  for  rent,  show  that  his 

92  89  X 

income  is  v?^A  under  the  old  tax  rate  and  -r-(A-|-— r^A)  under  the 

advanced  rate. 

Thus,^-Q^(A+^A)=^A. 

Divide  each  term  by  A  and  solve  the  equation  for  x. 

2.  A  contractor  needs  40,500  bricks  for  a  building.  His 
experience  has  shown  that  usually  3.5  per  cent  are  spoiled. 
How  many  bricks  must  he  order  ? 

3.  A  man  paid  $6,200  for  his  house.    His  tax  is  $77,  his  coal 

bill  is  $72,  and  he  spends  $50  a  year  for  repairs.     If  money  is 
worth  5  per  cent,  how  much  is  his  monthly  rental? 

4.  A  man  invested  $3,000,  part  at  5  per  cent  and  the 
remainder  at  6  per  cent,  obtaining  an  income  of  $157  per  year. 
How  much  has  he  invested  at  each  rate  ? 


216  SECOND-YEAR  MATHEMATICS 

333.  Loss  of  weight  problems. 

If  a  body  weighing  2  lb.  in  the  air  is  suspended  by  a  cord 
and  weighed  when  immersed  in  water,  it  will  weigh  less  than 
2  pounds.  It  can  be  shown  that  as  the  weight  of  the  water  the 
body  displaces  the  loss  of  weight  is  the  same. 

1.  A  mass  of  gold  weighs  97  oz.  in  air  and  92  oz.  in  water, 
and  a  mass  of  silver  weighs  21  oz.  in  air  and  19  oz.  in  water. 
How  many  ounces  of  gold  and  of  silver  are  there  in  a  mass  of 
gold  and  silver  that  weighs  320  oz.  in  air  and  298  oz.  in  water  ? 

Solution :  (1)  Let  x  be  the  number  of  ounces  of  gold  in  the  mass. 

(2)  Then  320 —x  is  the  number  of  ounces  of  silver. 

(3)  Since  97  ounces  of  gold  lose  5  ounces,  1  ounce 

loses  ^\  of  an  ounce. 

(4)  Since  21  ounces  of  silver  lose  3  ounces,  1  ounce 

loses  Jj-  of  an  ounce. 

(5)  Therefore  the  loss   of  x  ounces  of  gold  is  ^ 

ounces,  and  the  loss  of  320  —x  ounces  of  silver 

.    3(320 -a:) 
'^         21        * 

(6)  Then  |^+^^^^^  =22  is  the  loss  of  the  whole 

mass. 

(7)  The  root  of  this  equation  is  the  required  number. 

2.  A  pound  of  lead  loses  yV  of  a  pound,  and  a  pound  of  iron 
loses  yy  of  a  pound  when  weighed  in  water.  How  many  pounds 
of  lead  and  of  iron  are  there  in  a  mass  of  lead  and  iron  that  weighs 
159  lb.  in  air  and  143  lb.  in  water  ? 

3.  If  38  oz.  of  gold  lose  2  oz.  when  weighed  in  water  and  if 
30  oz.  of  silver  lose  3  oz.  when  weighed  in  water,  what  is  the 
amount  of  each  in  a  mass  of  gold  and  silver  that  weighs  106  oz. 
in  air  and  99  oz.  in  water  ? 

|4.  If  19j  lb.  of  gold  and  10|  lb.  of  silver  each  lose  one  pound 
when  weighed  in  water,  how  much  gold  and  silver  is  contained 
in  a  mass  of  gold  and  silver  that  weighs  20  lb.  in  air  and  18i  lb. 
m  water? 


FRACTIONS.  FRACTIONAL  EQUATIONS 


217 


Trigonometric  Relations 

334.  The  exercises  below  give  practice  in  the  opera- 
tions with  fractions. 


Prove  the 

following  trigonometric  identities: 

J. 

l+tan2 

1-  1 

cos"^  A 

A] 

aalysis: 

Assume 
l+tanM  = 

1 

C0S2 

A 

Tl 

len 

'+!■ 

1 

"62 

C2 

(See  Fig.  266.) 

C2 
62 

Why? 

Substituting  for  62+a2  its  equal  c^, 


Fig.  266 


c      c 

r-  =  z-,  which  is  an  identity. 

62    62 


Startin;r  from  the  statement 


62    62  ' 


by  reversing  the  steps 


of  the  analysis,  we  may  now  prove  that  l+tan2  A 


1 


cos2  A  ' 


In  exercises  2-18  reversing  the  steps  involves  no  par- 
ticular difficulties.  That  part  of  the  proof  may  therefore 
be  omitted. 


2.  cos  A 


sin  A 

tan  A 


-    .       .     cos  A     1 
3.  tan  A  •    .     ,  =1 


4. 


sin  A 
1 


1 


cos  A    tan  A     sin  A 
1 


6.  sin  A  • 
8. 


1 


tan  A 
1  1 


tan2  A     sin2  A 
sin  A + cos  A 


=cos  A 
-1 


6.  cos  A  •  tan  A 


sin  A 


^1 


t9. 


1+tanA 
1 


=  cos  A 


sin  A 


sin  A  =cos  A 


tan  A 


218  SECOND-YEAR  MATHEMATICS 


10.  -J-r  •-^i-r=tanA+     ^ 


cos  A    sin  A  tan  A 

I 

11.  >^1— sin^A  =sin  A  •  ,    ^  . 
tan  A 

Assume  V^l  —  sin^  A  =  sin  A 


tan  A 


Then  Ji-t  =  ^.^ 

\       c^     c    m 


c^        c 


/62 


or  -  =  ^ 

c     c 

12.  tan  A  •  cos  A^V  1 — cos^  A 

+13    14-      ^      —     ^ 

•*■         ^tanM     sin^A 

15.  (l+tan2A)(l-sin2A)=l 


+^ -       1      ,  ,       i  cos  A 

Jl6.  j+tanA=; 


cos  A  1— tan  A  cos  A 


17.  ^— sinA  •  tanA=cosA 

cos  A 


ti8.  r  1  +  1  ][!-  1  ] 

Lcos  A    sm  A  J  L       tan  A  J 

"LcosA     sinAJL      tan  A  J 


FRACTIONS.    FRACTIONAL  EQUATIONS  219 

Summary 

335.  The  chapter  has  reviewed  and  extended  the  laws 
of  the  operations  with  fractions,  i.e. : 

1.  Addition  and  subtraction  of  fractions  having  the 
same  denominator. 

2.  Addition  and  subtraction  of  fractions  having  diiGfer- 
ent  denominators. 

3.  Multiplication  of  fractions. 

4.  Division  of  fractions. 

5.  Reduction  of  complex  fractions. 

336.  Fractional  equations  are  solved  by  multi- 
plying each  term  by  the  least  common  multiple  of  the 
denominator,  and  then  reducing  each  term  to  the  simplest 
form. 

337.  A  number  of  trigonometric  identities  were  proved. 


CHAPTER  XIV 

INEQUALITIES 

338.  Review  and  extension  of  the  axioms  and 
theorems  of  inequality  previously  established. 

1.  A  line'Segment,  or  an  angle,  is  greater  than  any  part  of 
itself  (§6). 

This  axiom  is  to  be  applied  only  when  the  magnitudes 
and  their  parts  are  all  positive.  For,  let  the  segment  AC, 
Fig.  267,  be  considered  positive. 

Then    CB   is    negative    and    ]^ ^ ^ 

AC-\-CB  =  AB.    For  this  reason  j^jq  267 

AC  and  CB  may  be  called  parts 

of  AB.     One  of  these  parts,  AC,  is  greater  in  magnitude 

than  AB. 

2.  The  sums  obtained  by  adding  unequals  to  equals  are 
unequal  in  the  same  order  as  the  unequal  addends  (§  10). 

For  example,  8>3 

and  4  =  4 

Hence,  12  >7 

3.  The  sufns  obtained  by  adding  unequals  to  unequMs 
in  the  same  order  are  unequal  in  the  same  order  (§11). 

For  example,  9>2 

and  4>3 

Hence,  13  >  5 

4.  If  three  magnitudes  are  so  related  that  the  first  is 
greater  than  the  second  and  the  second  greater  than  the  third, 
the  first  is  greater  than  the  third. 

220 


INEQUALITIES  221 

For,  if  a>6  and  b>c,  then  a-\-b>b+c. 

Subtracting  b  from  both  sides,  a>c. 

In  obtaining  the  last  inequaUty  the  following  axiom  is  used: 

5.  If  equals  are  subtracted  from  unequals,  the  remainders 
are  unequal  in  the  same  order  as  the  unequal  minuends. 

For  example,  10>4 

and  3  =  3 

Hence,  7>1 

6.  The  differences  obtained  by  subtracting  unequals 
from  equals  are  unequal  in  the  order  opposite  to  that  of  the 
subtrahend  (§12). 

For  example,  12  =  12 

and  8>  2 

Hence,  ,  4<10 

7.  The  products  obtained  by  multiplying  unequals  by 
positive  equals  are  unequal  in  the  same  order  as  the  multi- 
plicands. 

For  example,  10  <  15 

2  =  2 
.-.  •20<30 

8.  The  products  obtained  by  multiplying  unequals  by 
negative  equals  are  unequal  in  the  order  opposite  to  that  of  the 
multiplicands. 

For  example,  12  <  15 

-3=-3 


-36>-45 


9.  The  quotients  obtained  by  dividing  unequals  by  posi- 
tive equals  are  unequal  in  the  same  order  as  the  dividends. 

For  example,  20  <30 

2=  2 

10<15 


222  SECOND-YEAR  MATHEMATICS 

10.  The  quotients  obtained  by  dividing  unequals  by 
negative  equals  are  unequal  in  the  order  opposite  to  that  of  the 
dividends. 

For  example,  50  >     40 

-  2=-  2 
-25<-20 

11.  The  shortest  distance  between  two  points  is  the 
straight  line-segment  joining  the  points  (§3). 

The  following  theorems  express  inequaHties: 

12.  The  sum  of  two  sides  of  a  triangle  is  greater  than  the 
third  side,  and  their  arithmetical  differ- 
ence is  less  than  the  third  side.  ,        ^   . 

a/  \^ 

The  first  part  of  this  theorem  follows 
directly  from  11. 

The  second  part  follows  from  5.  For, 
let  a+6>c.  Fig.  268. 

Then,  c<a-\-h. 

Subtracting  a  from  both  sides  c—a<b. 

Similarly,  show  that  b—a<c;  that  c—b<a. 

13.  The  shortest  distance  from  a  point  to  a  line  is  the 
perpendicular  from  the  point  to  the  line  (§  35).     Prove. 

14.  //  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  them  are  unequal,  the  greater  angle  lying  opposite 
the  greater  side  (§33).     Prove. 

15.  //  two  angles  of  a  triangle  are  unequal,  the  sides 
opposite  them  are  unequal,  the  greater  side  lying  opposite 
the  greater  angle  (§  34).     Prove. 

16.  Any  point  not  on  the  perpendicular  bisector  of  a 
line-segment  is  unequally  distant  from  the  endpoints  (§71). 
Prove. 


INEQUALITIES  223 

17.  The  perpendicular  bisector  of  a  line-segment  is  the 
locus  of  all  points  equidistant  from  the  endpoints  of  the 
segment  (§71).     Prove. 

18.  Any  point  not  on  the  bisector  of  an  angle  is  not 
equidistant  from  the  sides  of  the  angle  (§  72).     Prove. 

19.  The  bisector  of  an  angle  is  the  locus  of  all  points 
within  the  angle  equidistant  from  the  sides  (§  72).     Prove. 

Solution  of  problems  by  means  of  inequalities 

339.  Many  problems  lead  to  relations  expressed  as 
inequalities.  These  ineqvMities  may  then  be  solved  by 
using  the  axioms  of  inequality  in  the  same  way  as  equa- 
tions are  solved  by  using  the  axioms  of  equality.  The 
following  exercises  will  show  the  solution  of  problems  by 
means  of  inequalities: 

EXEECISES 

1.  Express  relations  which  hold 
between  the  sides  of  the  triangle  in 
Fig.    269. 

2.  For  what  values  oi  x  6h  the  rela-  Yig  269 
tions  in  exercise  1  hold  ? 

a;4-4+9>a:  +  12.  Why? 

.-.     13>12.  Why? 

.*.    x  =  any  value,  i.e.,  any  value  of  x  will  satisfy  the  in- 
equality. 

9+x+12>x+4.  Why? 

.-.    21  >4.  Why? 

.*.     x  =  any  value, 

x+12-\-x+4t>9 
2a;+16>9 
2a;>-7 
x>-Si 
.'.    any  value  of  x  greater  than  —  3§  will  satisfy  all  three 
inequalities.    Why  ? 


224 


SECOND-YEAR  MATHEMATICS 


3.  For  what  values  of  x  may  the  following  expressions 
represent  the  lengths  of  the  sides  a,  b,  and  c,  of  a  triangle  ? 


a 

x-5 

2x4-3 

x+5 

7 

2x 

b 

x+7 

2x+2 

S-x 

x-S 

5 

c 

16 

21 

1 

9 

4x-7 

4.  Two  sides  of  a  triangle  are  9  and  24  inches.  Between 
what  limits  must  the  third  side  be  ? 

Let  X  denote  the  third  side. 
Then  a;+9>24.        Why? 

a;+24>9.        Why? 

9+24  >x.  Why? 
Find  the  values  of  x  satisfying  all  three  inequaUties. 

5.  There  are  $50  in  the  treasury  of  a  club.  The  club  wants 
to  buy  furniture  costing  between  $80  and  $90.  How  much 
should  be  raised  ? 

Let  X  be  the  number  of  dollars  to  be  raised,  etc. 


JG.  A  twentieth-century  limited  train  wants  to  make  the 
distance  between  New  York  an^  Chicago  (1,000  miles  approxi- 
mately) in  less  than  20  hours.  During  the  first  five  hours  it 
goes  at  the  rate  of  45  miles  per  hour.  During  the  next  7  hours 
it  goes  at  the  rate  of  57  miles  per  hour.  How  fast  should 
it  go  thereafter  to  cover  the  distance  within  the  desired 
time  ? 

JT.  A's  record  average  speed  on  a  2-mile  run  is  6  miles  per 
hour,  and  B's  is  5f  miles.     How  many  feet 
can  A  afford  to  give  B  as  a  handicap  ? 

8.  Prove  that  the  diameter  of  a  circle 
is  longer  than  any  other  chord  of  that 
circle. 

Show  that  AB  =  CO+OD>CD,  Fig.  270.  Fig.  270 


INEQUALITIES 


225 


9.  Prove  the  following: 

(a)  The  distance  between  the 
centers  of  two  circles  which  lie 
entirely  outside  of  each  other  is 
greater  than  the  sum  of  the  radii, 
Fig.  271. 

(6)  The  distance  between  the 
centers  of  two  circles  touching  each 
other  externally  is  equal  to  the  sum 
of  the  radii,  Fig.  272. 

(c)  The  distance  between  the 
centers  of  two  intersecting  circles  is 
less  than  the  sum  of  the  radii,  but 
greater  than  the  difference.  Fig.  273. 

{d)  The  distance  between  the 
centers  of  two  circles  touching  each 
other  internally  is  equal  to  the 
difference  of  the  radii,  Fig.  274. 

(e)  The  distance  between  the 
centers  of  two  circles,  one  of  which 
Ues  entirely  within  the  other,  is  less 
than  the  difference  of  the  radii. 
Fig.  275. 

10.  Prove  that  an  exterior  angle 
of  a  triangle  is  greater  than  either  of 
the  remote  interior  angles. 

Use  §  26. 

11.  In  Fig.  276  prove  tnat  x  is 
greater  than  y. 

12.  Prove  that  the  sum  of  the 
diagonals  of  a  quadrilateral  is  less 
than  the  perimeter,  but  greater  than 
the  semi-perimeter. 


Fig.  271 


Fig.  272 


Fig.  273 


Fig.  274 


Fig.  275 


Fig.  276 


226 


SECOND-YEAR  MATHEMATICS 


13.  The  lengths  of  the  diagonals, 
Fig.  277,  are  denoted  by  5x-i-4  and 
4x— 31.  By  means  of  the  relations 
in  exercise  12,  determine  the  integral 
values  of  x. 

14.  The  line  joining  a  vertex  of 
a  triangle  to  the  midpoint  of  the 
opposite  side  is  a  median  of  the 
triangle. 

Prove  that  the  median  to  one 
side  of  a  triangle  is  less  than  one- 
half  of  the  sum  of  the  other  two 
sides. 

In  Fig.  278  extend  BD  making 
D£?  =  5I>  and  drawee. 

HhenBEKBC^-CE. 

VTOveCE=BA, 


Fig.  277 


Fig.  278 


Fig.  279 


16.  Two  towns  are  located  at 
A  and  B  respectively,  Fig.  279. 
Determine  a  point  P  on  the  edge  of 
a  river,  XY^  so  that  the  distances 
from  P  to  A  and  B  may  be  piped 
with  the  least  amount  of  pipe. 

Draw  AA'±XY  and  make  CA'  =  CA. 

Draw  BA'  meeting  Z7  at  P. 

P  is  the  required  point. 

Show  that  BP'A  >BPA,  P'  being  any  other  point  on  the  edge  of 
the  river. 

340.  Theorem:  If  two  oblique  line-segments  drawn 
to  a  line  from  a  point  on  the  perpendicular  to  the  line  have 
unequal  projections,  the  oblique 
line-segments  are  unequal.  -^J 

Let  E A ±BC,  and  AF> AD, 
Fig.  280. 

Prove  thsitEF>ED. 


INEQUALITIES 

Proof: 

Lay  off  AD'  =  AD  and  draw  ED' 

Then 

x>y 

Since  y  = 

=  90°,          .-.     a:>90° 

.-.     2<90° 

.-.     x>z 

:,EF>ED' 

I 

EF>ED 

227 


and 

341.  Theorem:  Two  unequal  oblique  line-segments 
drawn  to  a  line  from  a  point  on  a  perpendicular  to  the  line 
have  unequ/il  projections. 

Given  CB>CA,  CD±AB, 

Fig.  281. 

To  prove  that    DB>DA. 

Proof  (indirect  method) : 

1 .  Assume  DB  =  DA,  then  CB  =  CA.        Why  ? 
This  contradicts  the  hypothesis. 

.'.    The  assumption  is  wrong  and  DBt^DA* 

2.  Assume  DB<  DA. 
Then  show  that  CB<CA. 

This  is  impossible.  Why?  and  DB  is  not  less 
than  DA, 

3.  Since  DB  is  not  equal  to  DA  and  not  less  than  DA, 
it  follows  that  D5> DA. 

In  some  of  the  following  theorems  the  points  and  lines 
do  not  all  lie  in  the  same  plane.  Before  studying  th^ir 
proofs  select  points  and  lines  in  the  classroom  to  illus- 
trate the  figure  given  in  the  textbook.  If  the  practice  is 
followed  until  it  becomes  a  habit  it  will  add  greatly  to 
clearness  of  thought. 

*The  symbol,  9^,  means  "is  not  equal  to." 


228 


SECOND-YEAR  MATHEMATICS 


342.  Theorem:     Prove  that  the  perpendicular  is  the 
shortest  line  from  a  point  to  a  plane. 

Let  A  BCD,  Fig.  282,  repre- 
sent a  plane  and  E  be  any  point 
not  in  the  plane. 

Let  EF  be  perpendicular  to 
A  BCD  and  G  be  any  other  point 
than  Fin  A  BCD. 

Draw  EG. 

To  piOYe  thsii  EF <EG, 


Fig.  282 


Proof:  In  triangle  EFG  we  have  EF±FG. 

For,  a  Hne  perpendicular  to  a  plane  is  perpendicular 
to  any  line  in  the  plane  passing  through  the  foot  of  the 
perpendicular. 

.-.    EF<EG{^S5). 

343.  Distance  from  a  point  to  a  plane.  The  length 
of  the  perpendicular  from  a  point  to  a  plane  is  the  distance 
from  the  point  to  the  plane. 


344.  Theorem:  Oblique  lines  drawn  from  a  point  to 
a  plane,  meeting  the  plane  at  points  equidistant  from  the 
foot  of  the  perpendicular,  are  equal. 

Given  AB±CDEF  and  A  any 
point  on  AB,  Fig.  283.  ^ 

BG  =  BH.* 
To  prove    AG=AH  ^  g, 

F 

Proof:  Show  that 

AABG^AABH. 


Fig.  283 


*  BG  and  BH  are  the  projections  of  AG  and  AH  upon  the  plane 
CD^/^  (see  §§353-355). 


INEQUALITIES 


229 


345.  Theorem:  Oblique  lines  drawn  from  a  point  to  a 
plane  meeting  the  plane  at  points  unequally  distant  from 
the  foot  of  the  perpendicular  are  unequal,  the  more  remote 
being  the  greater. 

Given  plane  CDEF,  AB±CDEF  and  BH>BG, 
Fig.  284. 

To  prove  AH  >  AG.  a. 


Proof:   Lay  off  BG  on  BH, 
making  BK  =  BG. 

Then  AK  =  AG,        Why? 

AH>AK.        §340. 

AH>AG.        Why? 


Fig.  284 


346.  Theorem:  Equal  oblique  lines  drawn  from  a  point 
to  a  plane  meet  the  plane  at  points  equidistant  from  the 
foot  of  the  perpendicular.     Prove. 

347.  Theorem:  Of  two  unequal  oblique  lines  drawn 
from  a  point  to  a  plane  the  greater  meets  the  plane  at  the 
greater  distance  from  the  foot  of  the  perpendicular. 

Let        AH>AG,  Fig.  284. 

Lay  off  BK  =  BG.     Then  AK  =  AG  (§  346). 

.*.    AH>AK,  by  substitution. 

.-.    BH>BKi^SU). 

.\    BH>BG.        Why? 


EXERCISE 


Given  a  point  A  on  a  perpendicular  to  a  plane.    Find  the 
locus  of  points  in  that  plane  having  a  given  distance  from  A, 


230 


SECOND-YEAR  MATHEMATICS 


348.  Theorem:  If  from  a  'point  inside  a  triangle j  line- 
segments  are  drawn  to  the  endpoints  of  one  side,  the  sum  of 
these  line-segments  is  less  than  the  sum  of  the  other  two  sides. 

Given  AABC,  Fig.  285,  and  a 
point  P  inside  the  triangle. 
To  prove  that 

AP+PC<AB-\-BC. 

Proof:  Prolong  AP  until  it  inter- 
sects EC  Sit  some  point,  as  D.  Fig.  285 
We  now  have : 

AP+PD<AB-{-BD.       Why? 
PC<PD+DC.        Why? 
Adding,  AP+PD+PCkAB+BD+PD+DC. 

Subtracting  PD  from  both  sides 

AP+PC<AB+BD+DC. 
.-.    AP+PC<AB-\-BC.        Why? 


EXERCISES 

1.  Prove  that  the  sum  of  the  three 
line-segments  joining  a  point  inside  of  a 
triangle  with  the  vertices,  is  less  than  the 
perimeter  of  the  triangle  but  greater  than 
its  semi-perimeter,  Fig.  286. 

Use  §  338,  exercise  12,  and  §  348. 


Fig.  286 


2.  Determine  be- 
tween what  Hmits  x 
must  lie  in  Figs.  287 
and  288. 

Apply  exercise  1. 

What  values  could 
X  have  if  it  were  re- 
quired to  be  an  in- 
teger ? 


Fig.  287 


Fig.  288 


INEQUALITIES 


231 


Fig.  289 


E 

Fig.  290 


3.  Construct  a  triangle  ABC,  the  sides  a  and  b  and  the  angle 
A,  opposite  one  of  them,  being  given,  Fig.  289. 

Construction:  On  a 
line  of  indefinite  length, 
as  AB,  construct  an  angle 
equal  to  angle  A.  On 
one  side  of  this  angle,  as 
AC,  lay  off  AD  =  h. 

With  D  as  center  and 
radius  a  draw  a  circle. 
This  circle  will  either  in- 
tersect AB  in  two  points,  or  it  will  touch  AB,  or  it  will  not  meet  AB 
at  all. 

Discussion:  We  will  consider  the  case  where  ZA  is  acute. 

1.  li  a<h,  the  length  of  the  per- 
pendicular from  D  to  AB,  the  circle  will 
not  meet  AB,  and  there  is  no  triangle 
satisfying  the  given  conditions,  i.e.,  no 
solution  of  the  problem  exists,  Fig.  289. 

2.  If  a^h  the  circle  will  touch 
AB  and  there  is  one  solution  of  the 
problem,  i.e.,  AADE,  Fig.  290. 

3.  If  a>h,  and  <6,  the  circle 
will  intersect  AB  in  two  pomts  F  and 
F'.  There  are  two  solutions,  i.e., 
AADF  and  /\ADF',  Fig.  291. 

4.  If  a  is  equal  to  b  the 
circle  will  meet  AB  in  A  and  in 
another  point,  F.  There  is  one  solu- 
tion, i.e.,  AADF,  Fig.  292. 

5.  If  a>b,  the  circle  will  meet 
AB  in  two  points  F  and  F',  but  only 
AADF  satisfies  the  conditions  of  the 
problem.  Fig.  293. 

4.  Express  trigonometrically 
the  length  of  the  perpendicular, 
h,  in  terms  of  b  and  A,  i.e.,  show 
that  h  =  b  sin  A. 

Find  sin  A  from  the  right 
triangle  ADE  (see  §248). 


Fig.  291 


Fig.  293 


232 


SECOND-YEAR  MATHEMATICS 


349.  Theorem:  In  the  same  circle  or  in  equal  circles, 
unequal  chords  are  unequally  distant  from  the  center  of  the 
circle^  the  shorter  chord  lying  at  the  greater  distance;  and, 
conversely,  chords  unequally  distant  from  the  center  are 
unequal,  the  chord  at  the  greater  distance  being  the  shorter 
chord. 


Given  OP=  OQ,  Fig.  294. 

Chord  ^5> chord  DE,  PP'±AB,  QQ'±DE. 

To  prove  PP'<QQ'. 


Proof:  Place  QQ  on  OP,    > 
so  that  Q  falls  on  P,  DonB,  I 
and  chord  DE  in  the  position  \ 
BC;  then  Q'  will  take  a  posi- 
tion as  at  Q'\ 

Draw  P'Q". 

Y4^ 

AB>DE. 

Why? 

.*. 

AB>BC. 

PP'±AB. 

Why? 

P'B  =  ^AB. 

Why? 

QQ'IDE. 

Why? 

PQ"LBC. 

BQ"  =  lBC. 

Why? 

Then 

P'B>BQ". 

Why? 

.'. 

x>y. 

Why? 

Since 

x+z  =  y-\-u. 

Why? 

/. 

z<u. 

Why? 

/. 

PP'<PQ", 

Why? 

,*. 

PP'<QQ\ 

Why? 

Fig.  294 


INEQUALITIES  233 

Conversely,    given    OP=OQ,    Fig.  294,    PP'±AB; 
QQ'±DE;  PP'<QQ\ 

To  prove  that  A5  >!)£:. 

Proof:  Proceed  with  the  steps  of  the  foregoing  demon- 
stration in  the  opposite  order. 

EXERCISES 

1.  Triangles  are  to  be  constructed  with  the  following  parts: 


1.  6  =  145 

a  =178 

A  =  41° 

2.  a=     6 

6=     3.5 

A  =  63° 

3.  a  =  140 

6=170 

A  =  40° 

4.  6=  28 

a=  23 

A  =  65° 

Without  constructing  the  triangle,  tell  the  number  of  solu- 
tions in  each  case  by  comparing  the  lengths  of  a,  6,  and  h,  as 
found  by  the  formulas  in  exercise  4,  §  348. 

t2.  Construct  the  triangles  in  exercise  1  and  see  if  the  con- 
structions verify  the  results  obtained  from  the  formula. 

3.  Discuss  exercise  3,  §  348,  for  angle  A  a  right  angle;  for 
angle  A  an  obtuse  angle. 

4.  Prove  that,  in  the  same  circle,  a  side  of  a  regular  inscribed 
decagon  is  less  than  a  side  of  a  regular  inscribed  pentagon,  but 
that  the  side  of  the  decagon  is  greater  than  half  the  side  of  the 
regular  pentagon. 

6.  Show  that  the  greater  the  number  of  sides  of  a  regular 
inscribed  polygon,  the  shorter  is  the  length  of  one  of  its  sides. 

6.  Prove  that  the  distance  from  the  center  of  a  circle  to  a 
side  of  a  regular  inscribed  polygon  is  greater,  the  greater  the 
number  of  sides  of  the  polygon. 


234 


SECOND-YEAR  MATHEMATICS 


350.  Theorem:  //  two  sides  of  one  triangle  are  equal 
to  two  sides  of  another  triangle  but  the  angle  included  between 
the  two  sides  in  the  first  is  greater  than  the  angle  included 
by  the  corresponding  sides  in  the  second;  then  the  third  side 
in  the  first  triangle  is  greater  than  the  third  side  in  the  second. 

Given  AAJ5C  and  DEF,  Fig.  295. 

AB  =  DE;  BC  =  EF;   ZB>  ZE. 
To  prove  that         AODF. 

Proof;  Place  ADEF  on  A  ABC  so  that  DE  falls  on 
ABj  D  on  A,  E  on  B,  and  EF  on  the  same  side  of  AB 
as  BC.    Then  EF  must  fall  within  Z  ABC.        Why  ? 

For  the  position  of  F  there  are  three  possibilities. 

I.  F  falls  below  AC,  as  at  F\  Fig.  295. 


irhd^ 


Fig.  295 


Then 


a>b. 
b  =  c. 

a>c. 
c>d. 
a>d. 


Why? 
Why? 
Why? 
Why? 
Why? 


.-.      AOAF'^ndAODF.        Why? 
II.  F  falls  on  AC,  as  at  F'',  Fig.  296. 


INEQUALITIES 


235 


Then 


AOAF". 
AODF. 


Why? 
Why? 


III.  F  faUs  above  AC,  as  at  F"\  Fig.  297. 


Then 


AF' 


AF' 


and 


'-\-F"'B<AC-^CB.      Why? 
F"'B=  CB.      Why? 

Why? 
Why? 


<AC. 
DF<AC. 


351.  Theorem:  //  two  sides  of  one  triangle  are  equal 
to  two  sides  of  another  triangle,  the  third  side  of  the  first 
triangle  being  greater  than  the  third  side  of  the  second;  then 
the  angle  opposite  the  third  side  of  the  first  triangle  is  greater 
than  the  angle  opposite  the  third  side  of  the  second  triangle. 

Given  APQR  and  XYZ,  Fig.  298. 
PQ  =  XY;  QR=YZ;  PR>XZ. 
To  prove  that  ZQ>  ZY. 

Analysis:  If  Q  =  Fwhatis  known 
about  the  triangles,  about  PR  and 
XZf 

Hence,  can  Q=Y  if  PR>XZ, 
as  here  given? 

What  do  we  know  about  PR  and  XZiiQ<Yf     Why  ? 
Then,  is  Q<7,  if  PR>XZ,  as  here  given? 
How,  then,  must  angles  Q  and  Y  compare,  if  PR>XZ  f 
Give  full  proof,  using  the  indirect  method. 


236 


SECOND-YEAR  MATHEMATICS 


352.  Theorem:  In  the  same  circle  or  in  equal  circles, 
the  arcs  subtended  by  unequal  chords  are  unequal  in  the  same 
order  as  the  chords;  and,  conversely,  chords  subtending  un- 
equal arcs  are  unequal  in  the  same  order  as  the  arcs. 


Given  QA  =  OB,  Fig.  299. 
To  prove  arc  CD  >  arc  EF. 


CD>EF. 


Fig.  299 


Proof:  Draw  radii 
AC,    AD,    BE,    and    BF. 

Show  that 

ZCAD>ZEBFi^d51). 

Place  OB  on  OA,  so 
that  EB  falls  on  CA,  E  on 

C,  B  on  A,  and  F  on  the  same  side  of  C  as  D. 
Then  BF  must  come  between  AD  and  AC,  as  in  posi- 
tion AF'.     Why? 

Hence  EF  comes  in  the  position  CF\  and  F'  falls 
on  the  circle  between  C  and  D. 

Then,  arc  CF'  <  arc  CD.         Why  ? 

also,  arc  CF'  =  arc  EF.         Why  ? 

.-.     SLYcEFKsiYcCD.         Why? 

Conversely,  given       QA  =  QB,  Fig.  299,  CD>EF. 

To  prove  chord  CD  >  chord  EF. 

Proof:   Draw  radii  AC,  AD,  BF,  and  BE,  and  place 
OB  on  QA  so  that  EB  coincides  with  CA. 

Since  CD>EF,  the  point  F  will  fall  between  C  and 

D,  as  at  /^',  and  the  line  BF  will  come  on  the  same  side  of 
AD  as  AC,  as  in  position  AF\ 

Then,  we  have :        ZCAF'KZ  CAD.  Why  ? 

also,  Z  CAF'  =  Z  ^B/^.  Why  ? 

and,  _     _ZCAD>ZEBF.  Why? 

Show  that  CD>EF  (§350). 


INEQUALITIES 


237 


EXERCISES 


1.  The  length  of  the  chords  AB  and  BC,  Fig.  300,  being 
6x-14  and  4x+20,  respectively,  and  the  lines  PP'  and  PP" 
being  16  and  10,  determine  x  and  the  chords. 


P'B  =  Sx-7. 
P"B=2x-\-10. 


Then,  (3a:-7)2  +  162  =  P5l 


and 


We  have  P'B  =  3a;  -  7.  Why  ? 

Why? 
Why? 
i2x  +  10y  +  W  =  PB\  Why? 

.-.     (3a:-7)2  +  162  =  (2x  +  10)2  +  102 

9x2-42x+49+2o6  =  4a;2+40a;4-100  +  100 
5x2 -82x+ 105  =  0 

•      ^_82=t=y822-4.5  .  105" 
10 


Then 


Fig.  300 


x  = 

82  ±68 
=     10     = 

=  15, 

or[H] 

AB-- 

=  76. 

CB-- 

=  80. 

How  is  the  truth  of  the  theorem  in  §  349  illustrated  by  these 
answers  ?  , 


2.  The  length  of  the  lines  AB  and  BC,  PP'  and  PP" 
(Fig.  300)  being  denoted  by  h,  U,  di,  and  d-z,  respectively,  deter- 


h 

h 

dl 

dl 

1 

2 

ts 

u 

2a-7 

6 

x+S 

4i+14 

4a -14 

12 

x+5 
lot -2 

2 

6 
6 

1 

3m+4 
4 
3 

mine  the  unknown  number  in  each  of  the  following  cases.     In 
every  case  test  by  §  349. 


238 


SECOND-YEAR  MATHEMATICS 


Fig.  301 


Lines  and  Planes  in  Space 

353.  Projection  of  a  solid  upon  a  plane.  Imagine  a 
model  of  a  geometric  solid,  such  as  a  cube  made  of 
wire,  with  only  the  edges  and  corners  represented.  Sup- 
pose this  skeleton  cube  placed  between  a  small  light 
and  the  black- 
board (Fig.  301). 
A  shadow  of  the 
cube  will  appear  on 
the  board,  giving  a 
picture  containing 
all  the  important 
lines  and  points  of 
the  solid.  A  draw- 
ing of  this  shadow 
will  give  a  very 
good  idea  of  the 
form  of  the  cube, 
solid. 

By  removing  the  light  {center  of  projection)  far  enough, 
the  projecting  rays  become  nearly  parallel,  as  in  the  case 
when  the  sunlight  is  the  center  of  projection.  The  pro- 
jecting rays  may  be  perpendicular  or  oblique  to  the  plane 
of  the  blackboard. 

We  shall  consider  only  projections  obtained  by  pro- 
jecting rays  that  are  parallel  to  each  other  and  perpen- 
dicular to  the  plane  containing  the  projections. 

354.  Projection  of  a  point  upon  a  plane.  The  foot 
of  the  perpendicular  drawn  from  a  given  point  to  a  given 
plane  is  the  projection  of  the  point  on  the  plane. 

Choose  some  point  in  the  classroom  as  the  tip  of  a  gas 
jet,  the  corner  of  a  desk,  etc.,  and  tell  what  its  projections  are 
on  the  floor,  on  the  side  wall,  the  end  wall,  and  the  ceiling. 


The  shadow  is  the  projection  of  the 


INEQUALITIES 


239 


355.  Theorem:     The  projection  upon  a  plane,  of  a 
straight  line  not  perpendicular  to  the  plane,  is  a  straight  line. 

For,  all  projecting  rays,  AA',  a   b  c  ^ 

BB',  CC\  Fig.  302,  being  parallel, 
lie  in  a  plane  passing  through  AD, 

Hence,  the  projections  of  all         /   \     '     1_— J  ,    / 


N 


points  of  AD  lie  in  the  line  of  inter- 
section of  planes  AD'  and  MN. 


M' 


Fig.  302 


356.  Theorem:  The  projection  upon  a  plane,  of  a 
straight  line  perpendicular  to  the  plane,  is  a  point.    Why  ? 

357.  Theorem:  The  acute  angle  formed  by  a  given  line 
and  its  projection  upon  a  plane  is  smaller  than  the  angle 
which  it  makes  with  any  other  line  in  the  plane  passing  through 
the  point  of  intersection  of  the  given  line  and  the  plane. 

Given  line  AB  meeting  plane  P  at  B,  and  BA\  the 
projection  of  AB  upon  P.  Let  BC  be  any  other  line  in 
plane  P  passing  through  B,  Fig.  303. 

To  prove  that  AA'BA<A CBA. 

Proof:  On  BC  lay  off  BD  =  BA\ 
Then,  AB  =  AB 

BA'  =  BD 
and,  AA'<Ai)  (§342) 


Fig.  303 


/.A'BAkADBA  (§351). 


EXERCISES 

1.  Find  the  length  of  the  projection  of  AB,  Fig.  303,  in 
terms  of  AB  and  /.ABA'. 

2.  Find  the  length  of  the  projection  upon  a  plane  of  a  line 
10  ft.  long  and  making  an  angle  of  60°  with  the  plane. 

Use  the  formula  derived  in  exercise  1. 

3.  How  does  the  length  of  AB,  exercise  2,  compare  with  the 
length  of  the  projection? 


240  SECOND-YEAR  MATHEMATICS 

Summary 

358.  The  chapter  has  taught  the  meaning  of  the  follow- 
ing terms: 

distance  from    a   point   to  a      projection  of  a  point  upon  a 

plane  plane 

median  of  a  triangle  projection  of  a  segment  upon 

projection  of  a  solid  upon  a         a  plane 

plane 

359.  The  axioms  and  theorems  on  inequalities  studied 
in  the  preceding  chapters  were  reviewed  and  extended. 

360.  The  use  of  inequalities  in  the  solution  of  problems 
was  shown. 

361.  The  following  theorems  were  proved: 

1.  The  diameter  of  a  circle  is  larger  than  any  other  chord 
of  the  circle. 

2.  An  exterior  angle  of  a  triangle  is  greater  than  either 
of  the  remote  interior  angles. 

3.  If  two  oblique  line-segments  drawn  ta  a  line  from  a 
point  on  a  perpendicular  to  the  line  have  unequal  projections, 
the  oblique  line-segments  are  unequal. 

4.  Two  unequal  oblique  line-segments  drawn  to  a  line 
from  a  point  on  a  perpendicular  to  the  line  have  unequal 
projections. 

5.  If  from  a  point  inside  a  triangle,  line-segments  are 
drawn  to  the  endpoints  of  one  side  the  sum  of  these  line- 
segments  is  less  than  the  sum  of  the  other  two  sides. 

6.  In  the  same  or  in  equal  circles  unequal  chords  are 
unequally  distant  from  the  center,  the  shorter  chord  lying 
at  the  greater  distance;  and  the  converse  of  this  theorem. 


INEQUALITIES  241 

7.  //  two  sides  of  one  triangle  are  equal  to  two  sides  of 
another  triangle,  but  the  angle  included  between  the  two  sides 
of  the  first  is  greater  than  the  angle  included  between  the 
corresponding  sides  in  the  second,  then  the  third  side  in  the 
first  is  greater  than  the  third  side  in  the  second;  and  the 
converse  of  this  theorem. 

8.  In  the  same  or  equal  circles,  the  arcs  subtended  by 
unequal  chords  are  unequal  in  the  same  order  as  the  chords, 
and  the  converse  of  this  theorem. 

362.  The  points  and  lines  in  the  following  theorems 
do  not  all  lie  in  the  same  plane : 

1.  The  perpendicular  is  the  shortest  distance  from  a  point 
to  a  plane. 

2.  Oblique  lines  drawn  from  a  point  to  a  plane,  meeting 
the  plane  at  points  equidistant  from  the  foot  of  the  perpen- 
dicular, are  equal. 

3.  Oblique  lines  drawn  from  a  point  to  a  plane  meeting 
the  plane  at  points  unequally  distant  from  the  foot  of  the 
perpendicular  are  unequal,  the  more  remote  being  the  greater. 

4.  Equal  oblique  lines  drawn  from  a  point  to  a  plane 
meet  the  plane  at  points  equidistant  from  the  foot  of  the 
perpendicular. 

5.  Of  two  unequal  oblique  lines  drawn  from  a  point 
to  a  plane  the  greater  meets  the  plane  at  the  greater  distance 
from  the  foot  of  the  perpendicular. 

6.  The  projection  upon  a  plane  of  a  straight  line,  not 
perpendicular  to  the  plane,  is  a  straight  line. 

7.  The  projection  of  a  straight  line  perpendicular  to  the 
plane,  upon  a  plane,  is  a  point. 


242  SECOND-YEAR  MATHEMATICS 

8.  The  acute  angle  formed  hy  a  given  line  and  its  pro- 
jection upon  a  plane  is  smaller  than  the  angle  which  it  makes 
with  any  line  in  the  plane  passing  through  the  point  of  inter- 
section of  the  given  line  and  the  plane. 

The  following  construction  was  taught: 

363.  To  construct  a  triangle  ABC,  the  sides  a  and  b  and 
the  angle  A ,  opposite  one  of  them,  being  given. 


CHAPTER  XV* 


LINES  AND  PLANES  IN  SPACE.     DIEDRAL  ANGLES. 
THE  SPHERE 

364.  Theorem:  //  a  line  is  perpendicular  to  each  of 
two  intersecting  lines  it  is  perpendicular  to  the  plane  deter- 
mined by  these  lines. \ 

Given  line  AB,  Fig.  304,  inter- 
secting plane  P  at  C. 
ACLCD,ACA_CE. 
ToproveAC±P. 

Proof:  Let  CF  be  any  line  in  P 
passing  through  C. 

Draw  a  straight  line  DE  inter- 
secting CD,  CF,  and  CE. 

Draw  AD,  AF,  and  AE. 

Lay  off  CB  =  CA  and  draw  BD,  BF,  and  BE. 

Show  that  in  plane  ADB,  DC  is  the  perpendicular 
bisector  of  AB. 

Hence,  DA  =  DB.  For,  any  point  on  the  perpendicu- 
lar bisector  of  a  line-segment  is  equidistant  from  the 
endpoints. 

Similarly  show  that  EA  =  EB. 

Show  that  ADEA  ^  ADEB. 

Show  that  AADF^ABDF  (s.a.s.). 
.-.     AF  =  BF. 

*  This  chapter  may  be  omitted  if  it  seems  desirable  to  shorten 
the  course. 

t  Originated  by  Euclid,  simplified  by  Cauchy. 
243 


244 


SECOND-YEAR  MATHEMATICS 


Fig.  304 


Since  FA=FB  and  CA  =  CB,  it 
follows  that  FC  is  perpendicular  to 
AB.  For,  if  two  points  on  a  given 
line  are  equidistant  from  the  end- 
points  of  a  segment,  the  given  line 
is  a  perpendicular  bisector  of  the 
segment. 

Since  it  has  been  shown  that 
AB  is  perpendicular  to  CF,  which 
represents  any  line  in  P  passing 
through  C,  it  follows  that  AB  is  perpendicular  to  P. 

365.  Problem:  Through  a  given  point  in  a  given  line 
pass  a  plane  perpendicular  to  the  given  line. 

Construct  two  lines  perpendicular  to  the  given  line  at  the 
given  point.  Pass  a  plane  through  those  lines.  This  is  the 
required  plane.     Prove. 

366.  Theorem:  All  the  perpendiculars  to  a  given  line 
at  a  given  point  lie  in  a  plane  perpendicular  to  the  given  line 
at  that  point. 

Given  AB,  Fig.  305,  AC±AB,  AD±AB,  AE±AB. 
To  prove  that  Unes  AC,  AD,  AE 
lie  in  the  same  plane. 

Proof  (indirect  method) : 

Let  P  be  the  plane  determined 
by  AC  and  AD. 

Then  P± .45.     Why? 

Let  AE  represent  any  of  the 
lines   perpendicular  to   AB  at  A. 

Assume  that  A E  is  not  in  plane  P. 

Then  plane  Q,  determined  by  AE  and  AB  will  inter- 
sect plane  P  in  a  straight  line,  as  AF, 


FiQ.  305 


Courtesy  qf  Walter  Sargent 
S.  MARIA  DEL  FIORE— FLORENCE,  ITALY 


The  picture  above  illustrates  the  use  of  geometrical  forms  of 
architecture. 


LINES,  PLANES.     DIEDRAL  ANGLES.     SPHERE      245 

Then  in  plane  Q,         AF±AB.        Why  ? 
and  AE±AB.        Why? 

The  last  two  statements  contradict  the  theorem  that 
in  a  plane  (as  plane  PQ)  only  one  perpendicular  can  be 
drawn  to  a  given  line  at  a  given  point. 

Therefore  the  assumption  is  wrong  and  AE  Ues  in 
plane  P.  ^^  >^ 

367.  Theorem:  At  a  /fl^  point  J^a  given  line 
only  one  plane  can  he  constructed  perpendicular  to  the 
line. 

Show  that  this  follows  from  §  366. 

368.  Theorem:  From  a  given  point  outside  of  a  given 
line  one,  and  only  one,  plane  can  he  constructed  perpendicu- 
lar to  the  line. 


Given  line  AB,  Fig.  306,  and 

A 

point  C  not  on  AB. 

To  construct  a  plane  through  C 
perpendicular  to  AB. 

Construction:    Draw  CDLAB. 

/» 

/ 

/  •••-/ 

T>mwDE±AB. 

Construct  the  plane,  P,  deter- 

B 

mined  by  CD  and  DE. 

Fig.  306 

This    is    the    required    plane. 

Why? 

Moreover,  P  is  the  only  plane 

A 

perpendicular  to  A  5  from  C. 

// 

r^-^- 

For,  if  plane  Q,  Fig.  307,  be  also 

//      D 

~^-^^/  / 

perpendicular  to  AB,  intersecting 

^-i:'7/ 

AB  m  D\  then  CD'  and  CD  would 

both  be  perpendicular  to  AB,    This 

B 

is  impossible.    Why  ? 

F] 

Q.307 

246 


SECOND-YEAR  MATHEMATICS 


Fig.  308 


369.  Problem:  At  a  given  point  in  a  given  plane  con- 
struct a  perpendicular  to  the  plane. 

Given  point  A,  Fig.  308,  in 
plane  P. 

Required  to  construct  at  A  a 
line  perpendicular  to  plane  P. 

Construction:  Draw  BCin plane 
P  passing  through  A . 

Construct  plane  Q±BC  at  A,  intersecting  plane  P 
in  AD. 

In  plane  Q  construct  AE±AD. 

AE  is  the  required  perpendicular. 

To  prove  this,  show  that  AE±AD  and  AE±AB. 

370.  Theorem:    Only  one  line  can  be  constructed  per- 
pendicular to  a  given  plane  at  a  given  point. 

Given  AB±P,  Fig.  309. 

To  prove  that  AJ5  is  the  only 

line  perpendicular  to  P  at  A. 

Proof:   Assume  that  A 5  is  not 
the  only  perpendicular  to  P  at  A. 

Then  let  AC  be  another  perpen- 
dicular to  P  at  A. 

Pass  plane  Q  through  AB  and  AC  cutting  P  in  DE. 

Show  that  AB  and  AC  are  both  in  Q  and  perpendicular 
io  DE. 

This  is  impossible,  and  the  assumption  that  AP  is 
not  the  only  perpendicular  to  P  at  A  is  wrong. 

371.  Problem:    From  a  point  outside  of  a  plane  con- 
struct a  line  perpendicular  to  the  plane. 

Given  plane  P,  Fig.  310,  and  point  A,  not  in  P. 
To  construct  a  perpendicular  from  A  to  P. 


Fig.  309 


LINES,  PLANES.     DIEDRAL  ANGLES.     SPHERE       247 


Construction:  In  P  draw  a 
line,  as  BC. 

DrawADXBC. 
InPdraw2)^J.BC. 
.  T>T2iW  AF1.de. 
AF  is  the  required  line. 

Proof:  Draw  FG  any  line 
through  F  in  plane  P  meeting 
BC'mG. 

Extend  AF  making  FA'  =  FA. 

Draw  A'G,  A'D,  and  AG. 

Show  that  5C±  plane  ADF. 

Show  that  AD  =  A'D. 

Show  that  AADG^  AA'DG. 

.',    AG=^GA'.        Why? 

,*.    FG  is  a  perpendicular  bisector  of  A  A' 

Show  that  AP± P. 


V 

\ 

/. 

w/v 

A 

F^ 

/.Wb  / 

V 


Fig.  310 


Why? 


372.  Theorem:  From  a  given  point  outside  of  a  given 
plane  only  one  line  can  he  constructed  perpendicular  to  the 
plane. 

State  the  hypothesis  and  conclusion. 

Proof  (indirect  method) : 

Assume  that  AB,  Fig.  311,  is 
not  the  only  perpendicular  from 
A  to  P.  Let  AC  be  another 
perpendicular  from  A  to  P. 

Draw  plane  Q,  determined  by 
AB  and  AC,  intersecting  P  in  BC. 

In  plane  Q  both  AB  and  AC  are  perpendicular  to 

This  is  impossible. 

Hence,  the  assumption  is  wrong,  and  AP  is  the 
perpendicular  from  A  to  P. 


Fig.  311 


BC, 

only 


248 


SECOND-YEAR  MATHEMATICS 


373.  Theorem:  Lines  perpendicular  to  the  same  plane 
are  parallel. 

Given  lines  AB  and  CD  perpen- 
dicular to  plane  P. 
To  prove  AB  \\  CD. 

Proof:  Draw  BD. 
In  P  draw  EF±BD,  and  lay  off 
DE  =  DF,  •  Fig.  312 

BE  =  BF  (any  point  on  the  perpendicular  bisector  to  a  line- 
segment  is  equidistant  from  the  endppints). 

.-.     AE  =  AF.     (§344.) 

Show  that  AD  is  the  perpendicular  bisector  of  EF. 

Thus,  EF  is  perpendicular  to  DA,  DB,  and  DC. 

Therefore  DB,  DA,  and  DC  lie  in  the  same  plane. 
Why? 

.*.  AB  and  CD  lie  in  that  plane.  For,  if  two  points 
of  a  line  lie  in  a  plane  the  line  lies  wholly  in  that 
plane. 

Since  AB  and  CD  are  also  both  perpendicular  to  BD, 
it  follows  that  AB  II  CD. 


374.  Theorem:  If  one  of  two  parallel  lines  is  perpen- 
dicular to  a  plane,  the  other  is  perpendicular  to  the  same 
plane. 

Let  AB,  Fig.  313,  be  parallel  to  CD. 

Let  AB  be  perpendicular  to  P. 

If  CD  is  not  perpendicular  to  plane 
P,  Fig.  313,  we  may  draw  DC'±P. 

Then    DC  II  BA    and    DC  II  BA. 
Why?  / 

This  is  impossible.     Why  ?  ^ 

Complete  the  proof.  Fig.  313 


c'c 


LINES,  PLANES.     DIEDRAL  ANGLES.     SPHERE       249 


375.  Theorem:  Two  lines  parallel 
to  the  same  line  are  parallel  to  each 
other. 

Let  A  il  B  and  C  II  B,  Fig.  314. 
To  prove  A  II  C. 

Proof:  Draw  plane  P±B. 

Then,  A  ±  P  and  C  _L  P.         Why  ? 

.-.     AWC.        Why? 


Fig.  314 


376.  Theorem:   If  two  lines  are  parallel,  a  plane  con- 
taining one  of  them  and  not  the  other,  is  parallel  to  the  other. 

Given  AB  ||  CD,  Fig.  315, 
and  plane  P  containing  CD,  but 
not  AB. 

To  prove  AB  ||  P. 


Proof:    Suppose  AB  not 
parallel  to  P. 

Then  AP  must  meet  P  at  some  Fig.  315 

point  E,  if  far  enough  extended. 

Show  that  point  E  is  in  planes  P  and  Q. 

Then  E  must  be  on  their  intersection,  CD. 

Hence,  AB  and  CD  meet. 

This  contradicts  the  hypothesis  that  AB  II  CD  and 
the  assumption  that  AP  is  not  parallel 
to  P  is  wrong. 

377.  Theorem:    //  one  of  two 

parallel   planes  is   perpendicular  to  a 
line,  the  other  is  also. 

Given     plane     P  II  Q,    Fig.     316. 
Plane  P±AA\ 

To  prove  plane  Q±AA'.  Fig.  316 


250  SECOND-YEAR  MATHEMATICS 

Proof:  Through  AA^  pass  planes  R  and  S,  meeting 
P  in  AC  and  AD,  and  meeting  Q  in  A'C  and  A'D\ 

Then,  AC  II  A'C 
and  AD\\A'D\        Why? 

AA'  is  perpendicular  to  AC  and  AD.        Why? 

.*.     AA' is  perpendicular  to  A 'C  and  A 'D'.     Why? 

.-.    AA'±Q.        Why? 

378.  Theorem:    //  two  intersecting  lines  are  parallel 
to  a  given  plane,  their  plane  is  parallel  to  the  given  plane. 

Given  lines  A  B  and  A  C.  , 


AB  and  AC  are  parallel  to  plane  P.     A    ^ 
To  prove  Q  II  P. 


Proof:  Draw  A  A' ±  P.                        /    a'^ 
Draw  plane  R,  passing  through  AA^    ^ 


7 


y 


and  AC,  and  plane  S  passing  through  Fig.  317 

AA'  and  AB. 

Then,       AA'±A'B'  and  A'C.       Why  ? 

AC  II  A'C,  for,  if  AC  meets  A'C,  it  will 
meet  P. 
Likewise,    AP  II  A'B'. 

.'.      AA'±AP  and  AC.  Why? 

.-.      AA'±Q.  Why? 

Show   that    Q  II  P.      Use   indirect  method.      Apply 
§368. 

379.  Theorem:    If  two  angles  not  in  the  same  plane 
have  their  sides  parallel  and  running 
in  the  same  direction,  the  angles  are 
equal  and  their  planes  are  parallel. 


Given  angles  A,  A',  Fig.  318, 
such  that  AB  II  A'P',  AC  II  A'C. 

To  prove  A  A  =  A  A',  P  II  P'.  Fig.  318 


LINES,  PLANES.    DIEDRAL  ANGLES.    SPHERE      251 

Proof:  Draw  AA'.    Lay  off  AB  =  A'B\  AC=^A'C'. 
Draw  BC  and  B'C. 
Draw  CC  and  BB\ 

Since  AB  is  equal  and  parallel  to  A'B\  ABB' A'  is  a 
parallelogram  and  AA'  is  equal  and  parallel  to  BB'. 
Likewise,  A  A'  is  equal  and  parallel  to  CC. 
:.     CC  is  equal  and  parallel  to  BB' .        Why  ? 
/.     AAJ5C  ^  A  A'5'C'.        Why  ? 
/.     AA  =  AA'. 

P  is  parallel  to  A'C  and  A'5'  (§  376). 
/.    PI1F(§378). 


Diedral  Angles 

380.  Theorem:  All  plane  angles  of  a 
diedral  angle  are  equal. 

Show  that  the  sides  of  the  plane  angles  x  and 
y,  Fig.  319,  are  parallel. 
Then  apply  §  379. 

381.  Theorem:  Two  diedral  angles  are 
equal  if  their  plane  angles  are  equal.  Con- 
versely, if  two  diedral  angles  are  equal  their 
plane  angles  are  equal. 

Given  diedral  angles  BC 
and  B'C  and  their  plane 
angles  EFG^E'FV. 

To  prove  BC=^B'C'. 

Proof :  Place  diedral 
angle  BC  on  diedral  angle 
B'C,  making  AEFG 
coincide  with  E'FV.  This 
may  be  done  because 
AEFG  and  E'F'G'  are  equal. 


Fig.  320 


252 


SECOND-YEAR  MATHEMATICS 


Then  CF  must  coincide  with  C'F\        Why? 

.*.  Face  A  must  fall  on 
face  A'  and  face  D  on  face 
D'.    Why? 

Hence,  the  diedral 
angles  coincide  and  are 
equal. 

The  student  may  prove 
the  converse  theorem. 

A  number  of  theorems 
on  diedral  angles  are  analo- 
gous to  theorems  on  angles 

and  may  be  proved  in  the  same  way.     Some  of  these 
theorems  are  stated  in  the  following  exercises: 


Fig.  320 


EXERCISES 

Prove  the  following: 

1.  All  right  diedral  angles  are  equal. 

2.  The  sum  of  two  adjacent  diedral  angles  formed  by  two 
intersecting  planes  is  180°. 

3.  Vertical  diedral  angles  are  equal. 

4.  Diedral  angles  which  are  complements  or  supplements 
of  the  same  or  of  equal  diedral  angles  are  equal. 

6.  If  two  parallel  planes  are  cut  by  a  transversal  plane — 
The  alternate  interior  diedral  angles  are  equal. 
The  corresponding  diedral  angles  are  equal. 
The  interior  diedral  angles  on  the  same  side  are  supple- 
mentary. 

6.  State  and  prove  the  converse  of  exercise  5. 

7.  The  bisecting  planes  of  a  pair  of  vertical  diedrals,  are 
perpendicular. 


LINES,  PLANES.     DIEDRAL  ANGLES.    SPHERE       253 

382.  Theorem:   If  a  line  is  perpendicular  to  a  plane, 
every  plane  passing  through  this  line  is  perpendicular  to  the 

plane. 

bA 
Given  AB±  plane  P,  Fig.  321,  / 

and  plane  Q  any  plane  passing        / 

through  AB.  / 


L 


y 


To  prove  that  Q±P. 


Fig.  321 


Proof:    In  plane  P  draw  AC±DE,  the  intersection 
of  P  and  Q. 

BA±DE.        Why? 

.-.     ABAC  is  the  plane  angle  of  B-ED-C.     Why  ? 
*.•    BA±AC,  ZBAC  is  a  right  angle. 
.-.     Q±P.        Why? 


EXERCISE 

Show  that  through  a  Hne  perpendicular  to  a  given  plane  any 
number  of  planes  may  be  drawn  perpendicular  to  the  given 
plane. 

383,  Theorem:  //  two  planes  are  perpendicular  to 
each  other,  a  line  drawn  in  one  of  them  perpendicular  to 
the  intersection  is  perpendicular  to  the  other. 


Owen  P±Q,AB±CD. 
To  prove  that  AP±Q. 

Proof:  In  plane  Q  draw 
BE±CD. 

Then  /.ABE  is  the  plane 
angle  of  A-DC-E. 

.'.     A  ABE  is  a  right  angle. 

.♦.    AB±BE. 

.'.    AB±Q.        Why? 


A 


f 


a 
Fig.  322 


Why? 


254 


SECOND-YEAR  MATHEMATICS 


EXERCISES 

Prove  the  following: 

1.  //  two  planes  are  perpendicular  to  each  other,  a  line  per- 
pendicular to  one  of  them  at  a  point  of  the  intersection  must  lie 
in  the  other. 

Let  AB,  Fig.  323,  be  perpendicular  to  Q  and  let  P  be  perpen- 
dicular to  Q. 

Suppose  AB  does  not  lie  in  P.  Then 
CB  may  be  drawn  perpendicular  to  DE  in 
plane  P. 

C5J.Q(§383). 

But  AB±Q  at  the  same  point  B. 

This  is  impossible,  etc. 

2.  If  from  a  point  in  one  of  two  perpendicular  planes  a  line 
is  drawn  perpendicular  to  the  other  it  must  lie  in  the  first  plane. 

Use  the  indirect  method  of  proof. 

384.  Theorem:  If  a  plane  is  perpendicular  to  two 
planes  it  is  perpendicular  to  the  line  of  intersection. 

Given  plane  P,  Fig.  324,  per- 
pendicular to  Q  and  to  R. 

To  prove  P±  the  line  of  inter- 
section AB. 

Proof:  At  A,  the  point  common 
to  P,  Q,  and  R,  draw  a  line  perpen- 
dicular to  P. 

This  line  must  lie  in  plane  Q.        Why? 

For  the  same  reason  it  must  lie  in  plane  R. 

It  is  therefore  the  intersection  of  Q  and  R. 

Hence,  the  intersection  of  Q  and  i?  is  a  line  perpendicu- 
lar to  plane  P. 

How  could  this  theorem  be  applied  to  test  whether 
the  line  of  hinges  of  a  door  is  perpendicular  to  the  floor  of 
a  room,  using  only  a  carpenter's  square  ? 


Fig.  324 


LINES,  PLANES.    DIEDRAL  ANGLES.     SPHERE       255 

385.  Theorem:  Through  a  line  not  perpendicular  to 
a  given  plane,  one  plane  and  only  one  may  he  passed  perpen- 
dicular to  the  given  plane. 

Given  ABnot  ±  to  P,  Fig.  325.  ^i« 

To  prove  that  through  AB  one  a<^\     Q 

plane  may  be  drawn  perpendicular  to 

P  and  only  one. 


u 


Construction:   From  any  point  G  Fig.  325 

on  A 5  draw  CD:lB. 

Draw  the  plane  Q  determined  hy  AB  and  CD. 

This  is  the  required  plane. 

Prove  that  a  _L  P. 

Q  is  the  only  plane  through  AB  perpendicular  to  P. 
For  if  another  plane  could  be  passed  through  AB  per- 
pendicular to  P,  it  would  follow  that  P±AP,  the 
intersection  of  the  two  planes.  This  contradicts  the 
hypothesis. 

EXERCISES 

Prove  the  following: 

1.  A  plane  perpendicular  to  the  edge  of  a  diedral  angle  is 
perpendicular  to  the  faces. 

2.  Through  a  point  within  a  diedral  angle  a  plane  may  be 
passed  perpendicular  to  each  face. 

3.  If  three  lines  are  perpendicular  to  each  other  at  the  same 
point,  each  line  is  perpendicular  to  the  plane  determined  by  the 
other  two. 

The  Sphere  /"^T'^'^. 

386.  Sphere.    Center.    Radius.  /      /  j  ^^^^ 

Diameter.    A    sphere  is  a   solid  ^^^'"]j|    P))ffl||^ 

bounded  by  a  surface,  all  points  of  \"*~t--l-f7^M 

which  are  equidistant   from  a  point  \>.^.^^^^^ 

within  called  the  center,  Fig.  326.  ^^^^^^ 

A  line-segment  from  the  center  to  Ym.  326 
the  surface  of  the  sphere  is  a  radius, 
asOA. 


256 


SECOND-YEAR  MATHEMATICS 


Fig.  327 


A  diameter  is  a  segment  passing  through  the  center  and 
terminated  by  the  surface,  as  BC. 

A  sphere  may  be  produced  by  revolving  a  semicircle 
about  the  diameter. 

3.87.  Preliminary  theorems: 

1.  All  radii  of  the  same  sphere  are  equal. 

2.  All  diameters  of  the  same  sphere  are  equal, 

3.  The  radii  of  equal  spheres  are  equal. 

4.  Spheres  having  equal 
radii  are  equal. 

388.  Section  of  a 
sphere.  The  intersection 
of  a  plane  with  the  surface 
of  a  sphere  is  a  section  of 
the  sphere,  as  the  curve 
ABCD,  Fig.  327. 

389.  Theorem:    The  section  of  a  sphere  made 
plane  is  a  circle. 

Given  a  sphere  0  cut 
by  a  plane  P,  making  the 
section  ABC. 

To  prove  that  A  5  C  is  a 
circle. 

Proof:  Let  A  and  B  be 
any  two  points  on  the  sec- 
tion ABC. 

Draw  the  radii  OA  and  OB. 

Draw  Oi)±  plane  P. 

Draw  AD  and  DB. 

ThenZ)A=DP  (see  §  346). 

.*.     ABC  is  a  circle,  since  all  points  on  ABC  are 
distant  from  D. 


by  a 


Fig.  328 


equi- 


LINES,  PLANES.    DIEDRAL  ANGLES.    SPHERE       257 

390.  Great    circle.    Small    circle.    Poles.    Axis.    A 

section  made  by  a  plane  passing 
through  the  center  of  a  sphere  is  a 
great  circle,  as  ABC,  Fig.  329. 

A  section  whose  plane  does  not 
pass  through  the  center  is  a  small 
circle,  as  A'B'C\  Fig.  329. 

The  diameter  perpendicular  to  the  yig.  329 

plane  of  a  circle  of  a  sphere  is  the  axis 
of  the  circle  and  the  extremities  of  the  diameter  are  the 
poles  of  the  circle. 

EXERCISES 

1.  Find  the  area  of  a  plane  section  of  a  sphere  of  radius  10, 
which  passes  6  units  from  the  center.     (Board). 

Show  the  truth  of  the  following  theorems: 

2.  The  axis  of  a  circle  passes  through  the  center. 

3.  The  diameter  of  a  sphere  passing  through  the  center  of  a 
circle  is  perpendicular  to  the  plane  of  the  circle. 

*  4.  All  great  circles  of  a  sphere  are  equal. 

-'  5.  Two  great  circles  bisect  each  other. 

6.  Through  two  points  on  the  surface  of  a  sphere,  not  the  end- 
points  of  a  diameter,  only  one  great  circle  can  be  drawn. 

How  many  points  determine  a  plane  ? 

What  third  point  must  be  selected  to  determine  a  circle 
on  the  sphere  ? 

When  do  two  given  points  and  the  center  of  the  sphere  not 
determine  a  plane  ? 

7.  Every  great  circle  bisects  the  sphere. 

For,  the  two  portions  into  which  the  great  circle  divides  the 
surface  of  a  sphere  can  be  made  to  coincide,  as  all  points  on  the 
surface  of  the  sphere  are  equidistant  from  the  center. 


258 


SECOND-YEAR  MATHEMATICS 


391.  Spherical  distance  between  two 
points.  The  length  of  the  minor  arc  of  a 
great  circle  passing  through  two  points 
is  the  spherical  distance  between  them. 

Thus  ADB,  Fig.  330,  is  the  spherical  dis- 
tance between  A  and  B. 


Fig.  330 


392.  Theorem^  All  points  on  a  circle  of  a  sphere  are 
equidistant  from  its  poles. 

Given  two  points  A  and  B,  Fig.  331, 
on  the  circle  AB  oi  the  sphere  0;  P  and 
P'  the  poles  of  circle  AB.  ^ 

To  prove  that  PA  =  fs. 

•  Proof:  Let  the  axis  PP^  intersect  the 
plane  of  circle  AB  in  C. 

Then  C  is  the  center  of  circle  AB. 
Why? 

.-.     CA  =  CB.         Why? 

.-.     PA=PB.        Why? 
.-.     PA=PB.         Why? 

393.  Polar  distance.  The  spherical  distance  from 
the  nearer  of  the  poles  of  a  small  circle  to  any  point  on  the 
circle  is  the  polar  distance  of  the  circle. 

The  polar  distance  of  a  great  circle  is  the  spherical 
distance  to  either  pole. 

394.  Quadrant.  One-fourth  of  the  length  of  a  great 
circle  is  a  quadrant. 

395.  Theorem:  The  polar  distance  of  a  great  circle 
is  a  quadrant.     Prove. 


LINES,  PLANES.     DIEDRAL  ANGLES.     SPHERE       259 


396.  Theorem:  //  a  point  on  the  surface  of  a  sphere  is 
at  the  distance  of  a  quadrant  from  each  of  two  given  points 
on  the  surface,  it  is  a  pole  of  the  great  circle  passing  through 
the  given  points. 

Given  points  A,  5,  and  C  on  a 
sphere,  Fig.  332;  AB  =  si  quadrant; 
AC  =  a  quadrant;  BCD  a  great  circle 
arc. 

To  prove  that  A  is  a  pole  of  BCD. 

Analysis:  If  A  is  a  pole  of  arc  BC, 
what  can  be  said  of  diameter  AOE? 

How  can  we  show  that  A  0£^±  plane 
of  OOf 

How  large  are  angles  AOB  and  AOC  ?    Give  proof. 

397.  Theorem:  The  intersection  of  the  surfaces  of  two 
spheres  is  a  circle  whose  plane  is  perpendicular  to  the  line 
of  centers  of  the  spheres  and  whose  center  is  in  that  line. 

Let  the  two  intersecting  spheres 
be  generated  by  rotating  circles,  A 
and  jB,  Fig.  333,  about  the  center- 
line  A 5  as  an  axis. 

To  prove  that  the  spherical  sur- 
faces intersect  in  a  circle  whose 
center  is  in  A  5  and  whose  plane  is 
perpendicular  to  A  5. 

Proof:  Let  CD  be  the  common  chord  of  circles  A  and  B. 

Then  A 5  is  the  perpendicular  bisector  of  CD.    Why  ? 

As  the  plane  of  circles  A  and  B  revolves  about  AB, 
C  describes  the  Hne  common  to  the  two  spheres  thus 
generated. 

Line  CE  always  lies  in  the  plane  perpendicular  to 
ABoXE.    Why? 

.*.    The  path  of  C  is  a  circle  in  that  plane.     Why  ? 

EXERCISE 

Two  spheres,  whose  radii  are  12  inches  and  5  inches  respect- 
ively, have  their  centers  13  inches  apart.  Find  the  area  of  the 
circle  in  which  these  two  spheres  intersect.     (Harvard.) 


Fig.  333 


260 


SECOND-YEAR  MATHEMATICS 


398.  Tangent  line.  Tangent  plane.  If  the  surface  of 
a  sphere  and  a  line  (plane)  have  only  one  point  in  com- 
mon, the  Hne  (plane)  is  said  to  be  tangent  to  the  sphere. 

399.  Theorem;  A  plane  tangent  to  a  sphere  is  per- 
pendicular to  the  radius  at  the  point  of  contact. 


Given   sphere   A,   Fig.   334, 

^ \ 

and  plane  P  tangent  to  A. 

A.             \ 

To  prove  that  P±AB. 

K^                1 

Proof:  Let   C  be  any  point 
in  P,  not  B. 

/p                    B                 c/ 

Then    C   is   outside    of   the 

Fig.  334 

sphere.        Why  ? 

.*.    AC>  radius. 

Why? 

.-.    AOAB. 

Hence,  AB  is  the  shortest  distance  from  A  to  plane  P. 

Why? 

.-.     P±AB. 

Why? 

400.  Theorem:    A   plane  perpendicular  to  a    radius 
of  a  sphere  at  the  outer  extremity  is  tangent  to  the  sphere. 

To  prove  this,  reverse  the  order  of  steps  in  the  proof  of  the 
preceding  theorem. 


Summary 

401.  The  chapter  has  taught  the  meaning  of  the  follow- 
ing terms : 


sphere 

center 

radius 

diameter 

section  of  a  sphere 


great  circle 

small  circle 

poles 

axis  of  a  circle 

polar  distance 


spherical  distance  be- 
tween two  points 
quadrant 
tangent  line 
tangent  plane 


LINES,  PLANES.    DIEDRAL  ANGLES.    SPHERE      261 

402.  The  following  theorems  were  proved: 

1.  If  a  line  is  perpendicular  to  each  of  two  intersecting 
lines  it  is  perpendicular  to  the  plane  determined  by  these 
lines. 

2.  All  the  perpendiculars  to  a  given  line  at  a  given  point 
lie  in  a  plane  perpendicular  to  the  given  line  at  the  point, 

3.  Only  one  plane  can  he  constructed  perpendicular 
to  a  given  line  at  a  given  point. 

4.  Only  one  plane  can  he  constructed  perpendicular 
to  a  given  line  from  a  point  outside  of  the  line. 

5.  Only  one  line  can  he  constructed  perpendicular  to  a 
given  plane  at  a  given  point. 

6.  From  a  point  outside  of  a  given  plane  only  one  line 
can  6,e  constructed  perpendicular  to  the  plane. 

7.  Lines  perpendicular  to  a  plane  are  parallel. 

8.  //  one  of  two  parallel  lines  is  perpendicular  to  a 
plane,  the  other  is  perpendicular  to  the  same  plane. 

9.  Two  lines  parallel  to  the  same  line  are  parallel  to 
each  other. 

10.  If  two  lines  are  parallel,  a  plane  containing  one  of 
them  and  not  the  other,  is  parallel  to  the  other. 

11.  If  one  of  two  parallel  planes  is  perpendicular  to  a 
line  the  other  is  also. 

12.  If  two  intersecting  lines  are  parallel  to  a  given  plane, 
their  plane  is  parallel  to  the  given  plane. 

13.  If  two  angles  not  in  the  same  plane  have  their  sides 
parallel  and  running  in  the  same  direction,  the  angles  are 
equal  and  their  planes  are  parallel. 

14.  All  plane  angles  of  a  diedral  angle  are  equal. 


262  SECOND-YEAR  MATHEMATICS 

15.  //  two  diedral  angles  are  equal  their  plane  angles 
are  equal. 

16.  Two  diedral  angles  are  equal  if  the  plane  angles  are 
equal. 

17.  If  a  line  is  perpendicular  to  a  plane  every  plane 
through  this  line  is  perpendicular  to  the  plane. 

18.  //  two  planes  are  perpendicular  to  each  other  a  line 
drawn  in  one  of  them  perpendicular  to  the  intersection  is 
perpendicular  to  the  other. 

19.  //  two  planes  are  perpendicular  to  each  other  a  line 
perpendicular  to  one  of  them  at  a  point  of  the  intersection 
must  lie  in  the  other. 

20.  //  from  a  point  in  one  of  two  perpendicular  planes 
a  line  is  drawn  perpendicular  to  the  other,  it  must  lie  in  the 
first  plane. 

21.  If  a  plane  is  perpendicular  to  two  planes  it  is  per- 
pendicular to  their  intersection. 

22.  Through  a  line  not  perpendicular  to  a  given  plane, 
one  plane  and  only  one  may  he  passed  perpendicular  to  the 
given  plane. 

23.  The  section  of  a  sphere  made  by  a  plane  is  a  circle. 

24.  The  axis  of  a  circle  passes  through  'the  center. 

25.  The  diameter  of  a  sphere  passing  through  the  center 
of  a  circle  is  perpendicular  to  the  plane  of  the  circle. 

26.  All  great  circles  of  a  sphere  are  equal. 

27.  Every  great  circle  bisects  the  sphere. 

28.  Through  two  points  on  the  surface  of  a  sphere,  not  the 
endpoints  of  a  diameter,  only  one  great  circle  can  be  drawn. 

29.  All  points  on  a  circle  of  a  sphere  are  equidistant 
from  its  poles. 


LINES,  PLANES.     DIEDRAL  ANGLES.     SPHERE       263 

30.  The  polar  distance  of  a  great  circle  is  a  quadrant, 

31.  //  a  point  on  the  surface  of  a  sphere  is  at  the  dis- 
tance of  a  quadrant  from  each  of  two  given  points  on  the 
surface,  it  is  a  pole  of  the  great  circle  passing  through  the 
given  points. 

32.  The  intersection  of  two  spherical  surfaces  is  a  circle 
whose  plane  is  perpendicular  to  the  line  of  centers  of  the 
spheres,  and  whose  center  is  in  that  line. 

33.  A  plane  tangent  to  a  sphere  is  perpendicular  to  the 
radius  at  the  point  of  contact. 

34.  A  plane  perpendicular  to  a  radius  of  a  sphere  at  the 
outer  extremity  is  tangent  to  the  sphere. 

35.  To  determine  the  diameter  of  a  material  sphere. 
403.  The  following  constructions  were  taught: 

1.  Through  a  given  point  in  a  given  line  pass  a  plane 
perpendicular  to  a  given  line. 

2.  From  a  given  poinjfc  outside  of  a  given  line  construct 
a  plane  perpendicular  to  the  given  line. 

3.  At  a  given  point  in  a  given  plane  construct  a  per- 
pendicular to  the  plane. 

4.  From  a  point  outside  of  a  plane  construct  a  line 
perpendicular  to  the  plane. 

5.  To  pass  a  plane  perpendicular  to  a  given  plane,  that 
shall  contain  a  line  not  perpendicular  to  the  given  plane. 


CHAPTER  XVI 

LOCI.    CONCURRENT  LINES 

Loci 

404.  Locus.     When  a  point  moves  it  traces  a  path 
whose  shape  is  determined  by  the  conditions  under  which 
the  point  moves.     Thus,  a  stone 
falUng  from  rest  moves  along  a 
straight  line,  a  particle  projected 
obhquely  into  space  moves  along 
a  curve,  which  is  practically  a  -pio.  335 
parabola,  Fig.  335. 

In  the  study  of  geometry  we  have  learned  that  the 
location  of  all  points  in  a  plane  at  a  given  distance  from 
a  fixed  point  is  a  circle;  that  the  place  of  all  points  of  a 
plane  at  equal  distances  from  two  fixed  points  is  a  straight 
line,  the  perpendicular  bisector  of  the  segment  joining  the 
given  points. 

The  place  of  all  points  satisfying  some  specified 
condition  and  not  containing  other  points  is  called  the 
locus  of  the  points.  Locus*  is  a  Latin  word,  meaning 
^' place.'' 

405.  Determination  of  a  locus.  To  determine  the 
locus  of  a  point  mark  a  number  of  positions  of  the  point. 
From  these  points  it  will  be  possible  to  obtain  a  notion  of 
the  locus. 

Thus,  marking  several  positions  of  the  pedal  of  a  bicycle 
on  a  wall  beside  a  walk  suggests  the  locus  of  the  pedal. 

*  The  plural  of  locus  is  loci. 

264 


LOCI.    CONCURRENT  LINES 


266 


EXERCISES 

1.  A  circle  C,  Fig.  336,  is  rolled  without  sliding  along  the 
edge  of  a  ruler  AB.    Find  the  locus  of  a  point  P  on  the  circle. 

Cut   a   circle   from 
cardboard    and    roll    it  ^ 

carefully  along  the  ruler. 
By  pricking  through 
with  a  pin,  mark  a 
number  of  positions  of 
P.  Draw  a  smooth 
curve  through  the  points  thus  obtained 
cycloid. 


Fig.  336 
The  locus  of  P  is  called  a 


Fig.  337 


2.  Draw  two  perpendicular  lines, 
Fig.  337.  On  a  piece  of  tracing 
paper  draw  a  segment  AB  and  mark 
a  point  P  on  AB.  Move  AB  so 
that  B  sUdes  along  OY  and  A  along 
OX  and  mark  a  number  of  positions 
of  P.  Draw  the  locus  of  P.  The 
locus  will  be  a  quarter  of  an  ellipse. 

3.  What  is  the  locus  of  points  in  a  plane  having  a  given 
distance  from  a  given  line  ? 

Mark  several  points  at  the  given  distance  from  the  given  line. 
Their  position  will  suggest  the  locus. 

4.  What  is  the  locus  of  points  in  a  plane  at  equal  distances 
from  two  given  parallel  lines  ? 

6.  What  is  the  locus  of  points  in  space  having  a  given  dis- 
tance from  a  given  point  ? 

6.  What  is  the  locus  of  points  in  space  equally  distant  from 
two  given  points? 

7.  What  is  the  locus  of  points  in  space  equally  distant  from 
two  parallel  lines  ? 

8.  What  is  the  locus  of  points  in  space  having  a  given  dis- 
tance from  a  given  line  ? 


266 


SECOND-YEAR  MATHEMATICS 


9.  What  is  the  locus  of  points  in  space  at  equal  distances  from 
three  given  points?     (See  §  411.) 

406.  Proof  for  a  locus.  The  locus  of  points  satisfying 
given  conditions  must  contain  all  points  satisfying  these 
conditions  and  no  other  points,  i.e. : 

I.  Every  point  on  the  locus  must  satisfy  the  given  con- 
ditions. 

II.  (a)  Every  point  satisfying  the  conditions  must  lie  on 
the  locus,  or 
(h)  Any  point  not  on  the  locus  must  not  satisfy  the 
conditions. 

407.  Theorem:  The  locus  of  points  in  a  plane  equi- 
distant from  two  given  points  is  the  perpendicular  bisector 
of  the  segment  joining  these  points. 

Proof:  I.  Show  that  every  point  on  the  perpendicular 
bisector  is  equidistant  from  the  two 
points. 

II.  Let  PA  =  PB,  Fig. 
338.  Let  PC  be  a  line  drawn  from 
P  to  the  midpoint,  C,  of  AB. 

Show  that  x  =  2/.  Fig.  338 

408.  Theorem:  The  locus  of  points  in  a  plane  which 
are  within  an  angle  and  equidistant  from  its  sides  is  the 
bisector  of  the  angle. 

Proof:    I.  Show  that  every  point 
on   the  bisector  is  equidistant  from    a- 
the  sides. 

11.  If  PBIAB,   Fig.  339,  c 

PC±AC  and  BP  =  PC,  show  that  x  =  y.  Fig.  339 


LOCI.    CONCURRENT  LINES 


267 


409.  Theorem:    The  locus  of  points  in  a  plane  at  a 

given  distance  from  a  given  point  is  the  circle  whose  center 

is  the  given  point  and  whose  radius  is  equal  to  the  given 

distance. 

'P 

Proof:  I.  Every  point  on  the  circle,        y"^  ^^A 

Fig.  340,  has  the  given  distance  from  the     f  /  \^ 

given  point.     Why? 

II.  Show  that  a  point  P,  not 

on  the  circle,  is  not  at  the  given  distance 

from  the  given  point  C. 


Fig.  340 


410.  Theorem:    The  locus  of  points  in  a  plane  at  a 
given  distance  from  a  given  line 

consists   of  a  pair   of  lines      ' 1        ^~ 

parallel  to  the  given  line  and 
the  given  distance  from  it. 

Show  that  conditions  I  and 
II  are  satisfied  in  Fig.  341. 


'ZZJ. 


Fig.  341 


EXERCISES 

1.  Show  that  the  locus  of  the  centers  of 
all  circles  in  a  plane  tangent  to  a  given  Hne 
at  a  given  point  is  the  perpendicular  to  the 
given  line  at  that  point. 

2.  Show  that  the  locus  of  the  centers  of 
all  circles  in  the  same  plane  of  given  radius 
and  tangent  to  a  given  line  consists  of  two 
lines  parallel  to  the  given  line  and  at  the  given 
distance  from  it. 

3.  Show  that  the  locus  of  the  vertex  of  an 
angle  of  given  size,  x,  whose  sides  pass  through 
two  fixed  points  A  and  B  consists  of  two  arcs 

having  AB  as  chord  and  x  as  inscribed  angle.  (See  §  301  for  con- 
struction of  this  locus . )  Show  that  for  a  point  D,  Fig.  342,  outside 
of  the  circle  arc,  y  <x  and  for  a  point  E  within  the  circle  arc,  z>x. 


268 


SECOND-YEAR  MATHEMATICS 


4.  Construct  an  isosceles  triangle  having  given  the  base 
and  the  angle  opposite  the  base. 

5.  Find  the  locus  of  the  midpoints  of  parallel  chords  of  a 
circle. 

6.  Find  the  locus  of  the  midpoints  of  chords  of  a  circle 
equidistant  from  the  center. 

7.  Find  the  locus  of  the  midpoints  of  all  chords  passing 
through  a  given  point  on  the  circle,  Fig.  343. 

8.  Find  the  locus  of  the  centers  of  all 
circles   passing   through    two   given   points. 


9.  Find  the  locus  of  the  centers  of  all 
circles  tangent  to  a  given  circle  at  a  given 
point. 


Fig.  343 


10.  Find  the  locus  of  the  midpoints  of  all  segments  drawn 
from  one  vertex  of  a  triangle  and  terminated  by  the  opposite 
side. 

11.  Construct  a  circle  with  a  given  radius  which  shall  be 
tangent  to  each  of  two  intersecting  lines. 

411.*  Theorem:  The  locus  of  points  in  space  equi- 
distant from  all  points  on  a  circle  is  the  line  perpendicular 
to  the  plane  of  the  circle  at  the  center. 

Proof:  I.  Show  that  any  point  P 
on  the  perpendicular  at  C,  Fig.  344, 
is  equidistant  from  all  points  of  the 
circle.     (Use  §  344.) 

II.  Show   that  any  point  Fig.  344 

P'  not  on  the  perpendicular  at  C  is 
not  equidistant  from  all  points  of  the  circle.     (Use  §  345.) 


*  §§  411-413  may  be  omitted,  if  chapter  XV  bas  been  omitted. 


LOCI.    CONCURRENT  LINES 


269 


412.  Theorem:  The  locus  of  points  in  space  equi- 
distant from  two  given  points  is  the  plane  bisecting  the  seg- 
ment joining  these  points,  and  perpendicular  to  it. 

Proof:  I.  Show  that  any 
point  in  plane  P,  Fig.  345,  is 
equidistant  from  A  and  B. 

II.  Let  D  be  any  point 
not  in  plane  P,  and  let  DA  =  DB. 
Show    that    DC    is   perpen- 
dicular to  AB.    Hence,  DC  must 
lie  in  plane  P. 


^z^B 


Fig.  345 


413.  Theorem:  The  locus  of  a  point  within  a  diedral 
angle  and  equidistant  from  the  faces  is  the  plane  bisecting 
the  angle. 

Given  the  diedral 
angle  A-BC-D,  Fig. 
346.  Plane  P  bisects 
the  diedral  angle. 

To  prove  that  P 
is  the  locus  of  points 
equidistant  from  the 
faces  Q  and  R. 


Fig.  346 


Proof:   I.  Prove  that  any  point,  as  E,  in  plane  P,  is 
equidistant  from  Q  and  R,  as  follows : 

Draw  EF±Q  and  EH±R. 

Pass  plane  S  through  EF  and  EH. 

Then  S±Q,  and  S±R  (§  382). 
.-.    >SJ.B0C(§384). 

.-.     BO  is  perpendicular  to  FO,  EO,  and  HO.    Why  ? 

,'.  A  FOE  and  HOE  are  plane  angles  of  the  diedra] 
angles  formed  by  P  and  Q,  and  by  P  and  R.     Why  ? 


270 


SECOND-YEAR  MATHEMATICS 


.-.     ZFOE=ZHOE. 
Prove  AFOE^AHOE. 
Then  EF  =  EH. 


Why? 


^ 

p 

^ 

^'"P 

s 

^^ 

/ 

"^"-^ 

/ 

■"A.^ 

~A  !^- 

^ 

""^^^ 

> 

>:2^ 

11.  Prove  that  every  point  equidistant   from 
Q  and  R  lies  in  the 
bisecting  plane  P.  as 
follows: 

Prove  as  in  Case  I 
that  z^FO^  and /fO^ 
are  plane  angles  of 
diedral  angles  PQ 
and  PR. 

Since  it  is  given 
that  EF  =  EH,  we 
may  prove 

AFOE^AHOE  (hypotenuse  and  one  side). 

.-.     ZFOE=ZHOE. 

.'.     Diedral  angle  PQ  =  diedral  angle  PR. 

.'.     Plane  P  bisects  Q-BC-R. 

Hence,  P  is  the  required  locus. 


Fig.  346 


Concurrent  Lines 

414.  Median.  The  median  of  a  triangle  is  a  segment 
drawn  from  a  vertex  to  the  midpoint  of  the  opposite 
side. 

415.  Center  of  gravity  of  a  triangle.  From  cardboard 
cut  a  triangle.  Draw  the  three  medians  of  the  triangle. 
If  the  construction  is  made  carefully,  the  three  medians 
will  meet  in  a  point.  If  the  triangle  is  supported  by 
placing  a  pin  under  the  point  of  intersection,  the  triangle 
will  be  found  to  balance.  For  this  reason  the  point  of 
intersection  of  the  three  medians  of  a  triangle  is  called 
the  center  of  gravity  of  the  triangle. 


LOCI.    CONCURRENT  LINES  271 

416.  Concurrent  lines.  If  three  or  more  lines  pass 
through  the  same  point,  they  are  called  concurrent  lines. 

417.  Theorem:  The  medians  of  a  triangle  are  con- 
current in  a  point  which  lies  two-thirds  the  distance  from 
the  vertex  to  the  midpoint  of  the  opposite  side. 

Given  A  ABC,  Fig.  347,  with  the  medians  AE,  BF, 
and  CD. 

To  prove  that  A  E,  A-^^ 

BF,  and   CD  are  con-  /     \I^^^^-^-^^ 

current  and  that,  /\ii^^^^f~~--— Sl^-«^^ 

AO  =  lAE    A  A— ^^^^---V -^^"'^^''^-•••^^^  c 

BO  =  ^BF  Pj^  347 

CO  =  fCD. 

Proof:  AE  must  intersect  CD  at  some  point,  as  0. 
For,  if  A£'  does  not  intersect  CD,  it  follows  that 

AE  II  CD 

and   that    ZEAC+ Z  DC  A  =  1S0°.     Show    that    this    is 
impossible. 

Draw  KH  joining  K,  the  midpoint  of  A 0  to  H,  the 
midpoint  of  OC. 

Draw  DE,  DK,  and  EH. 

Then,  DE  ||  AC  and  DE  =  ^AC  (§  168,  exer- 

cise 2,  and  §  159,  exercise  2). 

Similarly,  KH  \\  AC  and  KH=^^AC. 

.'.     KHED  is  a  parallelogram  (§  125). 
EO  =  OK=KA. 
and,  DO  =  OH  =  HC. 

.'.       AO  =  fA£;  and  CO  =  f  CD. 

Similarly,  we  may  show  that  CD  and  BF  meet  in  a 
point  which  is  two-thirds  the  distance  from  B  to  F  and 
from  C  to  D,  i.e.,  at  0. 


272 


SECOND-YEAR  MATHEMATICS 


418.  Trisection  point.  The  two  points  dividing  a 
segment  into  three  equal  parts  are  trisection  points. 
Thus,  the  point  of  intersection  of  the  medians  of  a  triangle 
is  a  trisection  point  of  each  median. 

419.  Theorem:  The  perpendicular  bisectors  of  the 
sides  of  a  triangle  are  concurrent  in  a  point  equidistant 
from  the  vertices  of  the  triangle. 


Given  AABC,  Fig.  348,  and  DE, 
FG,  and  HK  the  perpendicular 
bisectors  of  AB,  EC,  and  CA, 
respectively. 

To  prove  that  DE,  FG,  and  HK 
are  concurrent  in  a  point  equidistant 

from  A,  B,  and  C. 

• 

Proof:  DrawDi^. 

ZEDB  =  90° 
A  GFB  =  90° 


Fig.  348 


.-.  ZEDB-i- ZGFB  =  1S0° 
,'.  ZEDF+ZGFD<180°.     Why?c 
.'.  DE  and  FG  must  intersect  at  Fig.  349 

some  point,  as  0,  Fig.  349. 

For,  if  DE  does  not  intersect  FG,  then 
DE  II  FG  and 
ZEDF+GFD  =  1S0°. 

OC  =  OB.       Why? 
OB  =  OA.       Why? 
.-.     OC  =  OA.       Why? 
.*.    HK  must  pass  through  0.     For,  the  perpendicular 
bisector  of  a  segment  is  the  locus  of  all  points  equidistant 
from  the  endpoints. 


LOCI.    CONCURRENT  LINES  273 

EXERCISES 

1.  Show  that  the  point  0,  Fig.  349,  is  the  center  of  the  cir- 
cumscribed circle  of  triangle  ABC. 

2.  Draw  the  circle  circumscribed  about  a  triangle. 

3.  Draw  a  circle  passing  through  three  points  not  in  the  same 
straight  line. 

420.  Circumcenter.  The  point  of  intersection  of  the 
perpendicular  bisectors  of  the  sides  of  a  triangle  is  the 
circumcenter  of  the  triangle. 

421.  Theorem:  The  bisectors  of  the  angles  of  a  triangle 
are  concurrent  in  a  point  which  is  equidistant  from  the  sides 
of  the  triangle. 


c 

^  „ 

c 

Fig.  350 

=^s 

'^            H 

Fig.  351 

Given  AABC,  Fig.  350,  with  AD,  BE,  and  CF,  the 
bisectors  of  A  A,  B,  and  C,  respectively. 

To  prove  that  AD,  BE,  and  CF  are  concurrent  in  a 
point  equidistant  from  AB,  BC,  and  CA. 

Proof:  Show  that  AD  and  BE  intersect,  as  at  0, 
Fig.  351. 

Draw  OH±AB,  OK±AC,  OL±BC. 

Then,  OH  =  OK.        Why? 

OH  =  OL.  Why? 
.-.  OK  =  OL.  Why? 
.*.     CF  must  pass  through  0.       Why  ? 


274 


SECOND-YEAR  MATHEMATICS 


EXERCISES 

1.  Show  that  the  point  0,  Fig.  351,  is  the  center  of  the  circle 
inscribed  in  triangle  ABC. 

2.  Inscribe  a  circle  in  a  triangle. 

422.  Theorem:    The  three  altitudes  of  a  triangle  are 
concurrent. 


Fig.  352 

Given  AABC,  Fig.  352,  with  AD1.BC,  BE±AC, 
and  CF±AB. 

To  prove  that  AD,  BE,  and  CF  are  concurrent. 

Proof;  Draw  B'C'±AD,  C'A'LBE,  and  A'B'LCF, 
forming  AA'B'C. 

Then,  AB  \\  A'B', 

BC  II  B'C,     . 

and  CAWC'A'.        Why? 

Show  that  5'C  =  A5  =  CA'. 

Hence,  CF  is  the  perpendicular  bisector  of  A'B'. 

Similarly,  show  that  AD  is  the  perpendicular  bisector 
of  A'C  and  that  BE  is  the  perpendicular  bisector  of  C'A'. 

.'.    AD,  BE,  and  CF  are  concurrent.        Why  ? 

423.  Orthocenter.  The  point  of  intersection  of  the 
three  altitudes  of  a  triangle  is  called  the  orthocenter  of  the 
triangle. 


LOCI.    CONCURRENT  LINES  275 

424.  Incenter.  The  point  of  intersection  of  the  bi- 
sectors of  the  interior  angles  of  a  triangle  is  called  the 
incenter  of  the  triangle. 


•     EXERCISE 

Show  that  the  bisectors  of  one 
interior  angle,  as  A,  Fig.  353,  and  of 
the  exterior  angles  at  B  and  C  are 
concurrent. 


425.  Excenter.     The  point  of 

intersection  of  the  bisectors  of  two  exterior  angles  of  a 
triangle  and  the  third  interior  angle  is  called  an  excenter 
of  the  triangle. 

1.  How  many  excenters  are  there  ? 

2.  Draw  a  triangle.  Construct  four  circles  tangent 
to  the  three  sides. 

3.  Prove  that  the  bisectors  of  the  angles  of  a  quadri- 
lateral circumscribed  about  a  circle  meet  at  a  point. 

426.  Historical  note.  The  ancients  even  before  Euclid's 
time  were  acquainted  with  the  theorems  of  the  medians,  of  the 
altitudes,  of  the  angle-bisectors  and  of  the  perpendicular  bisectors 
of  the  sides  of  a  triangle,  but  they  placed  no  great  importance 
upon  them.  They  used  the  incenter,  the  circumcenter ,  the  ortho- 
center,  and  the  center  of  gravity  in  constructions  but  they  did 
not  theorize  about  them.  Greek  mathematics  so  completely 
dominated  the  science  until  after  mediaeval  times  that  theorems 
not  given  by  EucUd  were  regarded  as  of  Httle  moment.  At  the 
beginning  of  the  eighteenth  century  the  neglected  theme  began 
to  be  studied.  In  1723  the  problem  was  raised,  to  construct 
a  triangle  having  given  the  position  of  its  center  of  gravity,  G, 


276 


second-yeXr  mathematics 


of  the  incenter,  I,  and  of  the  orthocenter,  0.     Nothing  worth 
mentioning  came  from  this  problem. 

In  1765  Euler  (1707-83)  attacked  and  solved  the  problem 
of  calculating  the  distance  of  the  points  0,  G,  and  I  from  one 
another  and  from  C,  the  circumcenter,  in  ternis  of  the  sides  a, 
b,  and  c.     He  found  that  OCG,  (see  figure) ,  is  a  straight  hne  and 


Altitude 

Angle  Bisector 

Median 


that  GC=^  GO.  The  straight  hne  OCG  was  later  named  in  his 
honor,  the  Eulerian  line.  In  1821  Poncelet  showed  that  the 
midpoints  of  the  sides,  the  feet  of  the  altitudes,  and  the  mid- 
points of  the  upper  segments  of  the  altitudes  of  a  triangle  all  he 
on  the  same  circle. 

In  1822  Feuerbach  (1800-1834)  also  discovered  this  circle. 
He  showed  that  its  center  M'  bisects  the  segment  CO,  and  that 
its  radius  equals  half  the  radius  of  the  circumscribed  circle 
(  =  r/2).  Germans  in  his  honor  call  this  circle  Feuerbach's 
circle  but  English  mathematicians  prefer  to  call  it  the  nine- 
point  circle. 

Feuerbach  also  showed  the  circle  to  be  tangent  internally  to 
the  inscribed  circle  and  externally  to  the  escribed  circle,  and  that 
the  segment  OG  of  the  Eulerian  line  is  divided  by  the  center 
M'  in  the  ratio  2:1.  Since  Feuerbach's  time  all  these  points 
and  properties  have  been  extensively  studied  from  varied  points 
of  view,   and   much   mathematical   knowledge   has  resulted. 


LOCI.    CONCURRENT  LINES  277 

Feuerbach's  circle  was  first  given  place  in  an  elementary  book 
on  geometry  by  C.  F.  A.  Jacobi  in  1834.  (See  Tropfke,  Ge- 
schichte  der  Elementar-Mathematik,  II.  Bd.,  S.  88-90.) 

Summary 

427.  The  chapter  has  taught  the  meaning  of  the  follow- 
ing terms: 

locus  center  of  gravity  of  circimicenter 

cycloid  a  triangle  incenter 

ellipse  concurrent  lines  excenter 

median  trisection  point  orthocenter 

428.  The  proof  for  a  locus  consists  in  showing — 

I.  That  every  point  on  the  locus  satisfiies  given  con- 
ditions. 

II.  (a)  That  every  point  satisfying  these  conditions 
lies  on  the  locus,  or 
(6)  That  every  point  not  on  the  locus  does  not 
satisfy  these  conditions. 

429.  The  following  theorems  were  proved: 

1.  The  locus  of  points  in  a  plane  equidistant  from  two 
given  points  is  the  perpendicular  bisector  of  the  segment 
joining  these  points. 

2.  The  locus  of  points  in  a  plane  which  are  within  an 
angle  and  equidistant  from  its  sides  is  the  bisector  of  the 
angle. 

3.  The  locus  of  points  in  a  plane  at  a  given  distance  from 
a  given  point  is  the  circle  whose  center  is  the  given  point  and 
whose  radium  is  equal  to  the  given  distance. 

4.  The  locus  of  points  in  a  plane  at  a  given  distance  from 
a  given  line  consists  of  a  pair  of  lines  parallel  to  the  given 
line  and  the  given  distance  from  it. 


278  SECOND-YEAR  MATHEMATICS 

5.  The  locus  of  points  in  space  equidistant  from  all 
points  on  a  circle  is  the  line  perpendicular  to  the  plane  of  the 
circle  at  the  center. 

6.  The  locus  of  points  in  space  equidistant  from  two 
given  points  is  the  plane  bisecting  the  segment  joining  these 
points  and  perpendicular  to  it. 

7.  The  locus  of  points  within  a  diedral  angle  equidistant 
from  the  faces  is  the  plane  bisecting  the  angle. 

8.  The  medians  of  a  triangle  are  concurrent. 

9.  The  perpendicular  bisectors  of  the  sides  of  a  triangle 
are  concurrent  in  a  point  equidistant  from  the  vertices  of  the 
triangle. 

10.  The  bisectors  of  the  angles  of  a  triangle  are  concurrent 
in  a  point  which  is  equidistant  from  the  sides  of  the  triangle. 

11.  The  three  altitudes  of  a  triangle  are  concurrent. 


CHAPTER  XVII 

REGULAR  POLYGONS  INSCRIBED  IN,  AND  CIRCUM- 
SCRIBED ABOUT,  THE  CIRCLE.    LENGTH 
OF  THE  CIRCLE 

Construction  of  Regular  Polygons 

430.  Regular  polygon.  A  polygon  that  is  both  equi- 
lateral and  equiangular  is  a  regular  polygon. 

431.  Regular  polygons  in  designs.  Regular  polygons 
are  involved  in  many  forms  of  decorative  design.  We 
use  them  in  the  tile  floor,  Fig.  354;   in  the  ornamental 


Fig.  354 


Fig.  355 


Fig.  356 


Fig.  357 


Fig.  358 


o 


NTTTs 


o 


N^^N    / 


/_sjk'_\ 


K^/1K^/I 


O 


O 


Fig.  359 


window,  Fig.  355;  in  linoleum  patterns,  Figs.  356-357; 
in  paper  doilies.  Fig.  358;  in  ceiling  panels.  Fig.  359, 
floor  borders,  furniture  designs,  etc. 

279 


280  SECOND-YEAR  MATHEMATICS 

Point  out  the  regular  polygons  in  Figs.  354-359. 
It  is  the  purpose  of  the  first  part  of  the  chapter  to 
learn  how  to  construct  regular  polygons. 

EXERCISES 

1.  Show  that  an  equilateral  triangle  is  a  regular  polygon. 

2.  Draw  a  quadilateral  that  is  equilateral  but  not  equi- 
angular.   What  is  such  a  quadrilateral  called  ? 

3.  Draw  an  equiangular  quadrilateral.  What  is  such  a 
quadrilateral  called? 

4.  Draw  a  quadrilateral  that  is  not  equiangular  and  not 
equilateral. 

6.  Show  that  a  square  is  a  regular  polygon. 

6.  Make  a  sketch  of  a  regular  pentagon;  hexagon;  octagon 
(8-side). 

432.  Inscribed  polygon.  A  polygon  whose  vertices 
lie  on  a  circle  is  an  inscribed  polygon.  The  circle  is  said 
to  be  circumscribed  about  the  polygon. 

Draw  an  inscribed  pentagon;  hexagon. 

433.  Circumscribed  polygon.  A  polygon  whose  sides 
are  tangent  to  a  circle  is  a  circumscribed  polygon.  The 
circle  is  said  to  be  inscribed  in  the  polygon. 

Draw  a  circumscribed  polygon. 

434.  The  theorems  in  §§  435  and  437  will  be  used  when 
we  wish  to  prove  that  an  inscribed  or  circumscribed  poly- 
gon is  a  regular  polygon.  They  show  that  the  construc- 
tion of  regular  inscribed  and  circumscribed  polygons 
depends  upon  the  problem  of  dividing  a  circle  into  a  given 
number  of  equal  parts. 


HOUSE  IN  NUREMBERG,  GERMANY 


TOWN  HALL.  WERNIGERODE.  GERMANY 


Write  an  essay  on  the  uses  of  mathematical  forms  in  artistic 
buildings,  using  the  pictures  in  this  book  as  illustrations. 


REGULAR  POLYGONS  AND  THE  CIRCLE         281 

435.  Theorem:  If  a  circle  is  divided  into  equal  arcs, 
the  chords  subtending  these  arcs  form  a  regular  inscribed 
polygon. 


Given  the  circle  0,  Fig.  360,  divided  into  equal  arcs, 
AB,  BC,  CD,  etc. 

The  polygon  ABCD  ....  formed  by  the  chords 
subtending  these  arcs'. 

To  prove  that  ABCD  ....  is  a  regular  inscribed 
polygon. 

Proof:  I.  Show  that  chords  AB,  BC,  CD,  ...  , 
are  equal. 

II.  In   triangles   ABC  and  EDC  show  that 
x=^y,  m  =  n  (^29S). 

.'.   ZD=ZB.        Why? 

Similarly,  prove  that  the  other  angles  of  the  polygon 
are  equal. 

Hence,  ABCD  ....  is  a  regular  inscribed  polygon. 
Why? 

436.  Theorem:  If  the  midpoints  of  the  arcs  sub- 
tended  by  the  sides  of  a  regular  inscribed  polygon  of  n  sides 
are  joined  to  the  adjacent  vertices  of  the  polygon,  a  regular 
inscribed  polygon  of  2n  sides  is  formed.     Prove. 


282 


SECOND-YEAR  MATHEMATICS 


437:  Theorem:  If  a  circle  is  divided  into  equal  arcs, 
the  tangents  drawn  at  the  points  of  division  form  a  regular 
circumscribed  polygon. 

Given  circle  0,  Fig.  361; 
PQ  =  QR=RS,  etc.;  AB,  BC,  CD, 
etc.,  tangent  to  circle  0,  forming  the 
circumscribed  polygon  ABCD  ....    e 

To  prove  A  B  CD  ....  a  regular  ^ 

polygon. 

Proof:  Draw  PQ,  QR,  RS,  .  ,  .  .  , 

etc. 

Prove  APBQ,  QCR,  RDS,  etc.,  congruent  isosceles 
triangles. 

ZA=ZB=  ZC,  etc. 

Since  AP  =  BQ,         Why? 

and  PB  =  QC,         Why? 

AB  =  BC.         Why? 

Similarly,  prove       BC  =  CD  =  DE,   etc. 

Hence,   ABCD    is    a    regular    polygon. 


438.  Theorem:  If  tangents  are 
drawn  to  a  circle  at  the  midpoints  of  the 
arcs  terminated  by  consecutive  points  of 
contact  of  the  sides  of  a  regular  circum- 
scribed polygon  a  regular  circumscribed 
polygon  is  formed  having  double  the 
number  of  sides.   Prove.    (See  Fig.  362.) 


Fig.  362 


EXERCISES 

1.  Prove  that  an  equilateral  inscribed  polygon  is  regular. 


Show  that  the  circle  is  divided  into  equal  arcs. 
§435. 


Then  apply 


REGULAR  POLYGONS  AND  THE  CIRCLE         283 


2.  Prove  that  an  equiangular  circumscribed  polygon  is 
regular. 

Show  that  the  circle  is  divided  into  equal  arcs.  Then  use 
§437. 

439.  Problem:    To  inscribe  a  square  in  a  given  circle. 

Given  circle  0,  Fig.  363. 
Required  to  inscribe  a  square 
in  circle  0. 

Analysis:  Since  the  square  is  a  ^ 
regular   quadrilateral,  we    can   in- 
scribe a  square  if  we  can  divide  the 
circle  into  four  equal  arcs. 

A   circle  may  be  divided  into  yiq.  363 

four   equal    arcs   by   dividing  the 
plane  around  the  center  into  four  equal  angles. 

Since  the  sum  of  the  angles  around  0  is  360°,  each  of 
the  four  equal  angles  must  be  90°. 

State  a  way  of  constructing  four  right  angles  at  0. 

Construction:   Draw  the  diameter  AB. 
Draw  diameter  CD±AB. 
Draw  AD,  DB,  BC,  and  CA. 
Then  ADBC  is  the  required  square. 

Proof:  a=6=c=ci=90°.    Why? 

.-.    Ab  =  DB  =  BC  =  CA.        Why? 
.*.    ADBC  is  a  regular  quadrilateral,  i.e.,  a  square. 
Why? 

440.  Problem:  To  circumscribe  a  square  about  a  given 
circle. 

Proceed  as  in  the  construction  in  §  439  and  draw  tangents 
at  ^,  B,  C,  and  D. 


284  SECOND-YEAR  MATHEMATICS 

EXERCISES 

1.  Denoting  the  side  of  the  inscribed  square  by  a,  the 
radius  by  r,  prove  that  a=rV2. 

The    problem    may    be    solved    by  ^.---^sr-^^ 

algebra,  or  by  trigonometry:  /y^'    X^v 

(a)  Apply  the  theorem  of  Pythagoras         /  y^      !      \.  \ 
to  the  sides  of  triangle  AOD,  Fig.  364.        //    ^       j ^       \^ 

(6)  Find  the  required  relation  using     ^K~ 
the  sine  of  45°.  \\ 

Notice    that    the    equation   a  =  rV2  \"  \ 

expresses  the  fact  that  the  side  of  the  ''^^-^ 

inscribed    square    varies    directly    as    the  ^ 

radius.  Yiq.  364 

Show  that  a  is  a  function  of  r. 

2.  Express  the  side  a  of  the  circumscribed  square  in  terms  of 
the  radius  r. 

3.  Express  the  perimeters  of  the  inscribed  and  circumscribed 
squares  in  terms  of  the  radius;  in  terms  of  the  diameter. 

4.  Prove  that  the  point  of  intersection  of  the  diagonals  of  a 
square  is  the  center  of  the  inscribed  and  circumscribed  circles. 

5.  Show  how  to  construct  regular  polygons  of  8,  16,  32,  etc., 
sides. 

6.  Show  that  the  number  of  sides  of  the  polygons  in  exer- 
cise 5  is  expressed  by  the  formula  2«,  where  n  is  a  positive  integer 
equal  to,  or  greater  than  2. 

441.  Problem:  To  inscribe  a  regular  hexagon  in  a 
given  circle. 

Analysis:  Into  how  many  equal  arcs  must  the  circle 
be  divided  ? 

How  large  must  the  central  angles  be  that  intercept 
these  arcs  ? 

State  a  simple  way  of  constructing  an  angle  of  60°. 


REGULAR  POLYGONS  AND  THE  CIRCLE 


285 


Construction:  With  A  as  center 
and  radius  AO,  Fig.  365,  draw  an  arc 
cutting  the  circle  at  J5. 

With  B  as  center  and  the  same 
radius  draw  the  arc  at  C.  \     ^ 

Similarly,    draw    arcs    at    D,    E,  f 
and  F. 

DrsLwihepolygoii ABCDEF.    This 
is  the  required  hexagon. 

Proof:  Draw  OA,  OB,  OC,  etc. 
Prove    that   a  =  h  =  c  =  d  =  e=f=QO°. 
Prove  ihsit  AB  =  BC. .......  =FA. 

Then  polygon  ABCDEF  is  regular. 


442.  Problem: 

about  a  given  circle. 


Why? 

To   circumscribe   a   regular   hexagon 


EXERCISES 

1.  Express  the  relation  between  the  side  a  of  the  regular 
inscribed  hexagon  and  the  radius  r. 

2.  Express  in  terms  of  the  radius  the  side  of  the  regular 
circumscribed  hexagon. 

Draw  OA  and  OK,  Fig.  366. 

Show  that  triangle  AOK  is  a  60°-30° 
light  triangle. 

HenceA0  =  2  ■  AK  =  a. 

Find  the  required  relation  between  a 
and  r, 

First  by  using  the  theorem  of  Pythagoras; 

Secondly,  by  ^.^ing  the  tangent  of  30''. 

Show  that  the  side  of  the  regular  circum- 
scribed hexagon  varies  directly  as  the  radius. 

Show  that  the  side  is  a  function  of  the  radius. 

3.  Inscribe    and   circumscribe   an   equilateral   triangle,   a 
regular  12-side,  24:-side,  etc. 


FiQ.  366 


286 


SECOND-YEAR  MATHEMATICS 


4.  Show  that  the  number  of  sides  of  the  polygons  in  exer- 
cise 3  are  given  by  the  formula  3  •  2^,  n  being  a  positive  integer, 
or  zero.     (See  exercise  7  below  for  value  of  2°.) 


Qh  q7  fiin 

6.  Show  that  —^d^;  —  =  a^;   —  =  a"»-",  m  being  greater 
a^  a^  a^ 


than  n,  and  m  and  n  being  positive  integers. 


fl"» 


6.  Show  that  ^  =  1;   -=1;  ^  =  1. 


7.  Assuming  that  —  =  a"*  ~  "  when  m  =  n;  show  that  —  =  a° . 

So  far  we  have  not  defined  the  expression  a°.    To  make  the 
results  of  exercises  6  and  7  agree,  we  shall  define  a^  to  mean  1. 

8.  Give  the  values  of  2°,  3°,  a:",  (a+6)o,  {2x-y-{-z)\ 

9.  Express  in  terms  of  the  radius  r,  the  side  of  the  inscribed 
equilateral  triangle. 

Show  that  OK,  Fig.  367  is  ^  (§  see  exercise  2). 

Obtain  the  required  relation  first,  by  using  the  theorem  of 
Pythagoras;  secondly,  by  using  the  tangent  of  60°. 
Express  your  result  in  the  language  of  variation. 


Fig.  367 


Fig.  368 


10.  Show  that  the  side  of  the  circumscribed  equilateral 
triangle  is  2rV I  (Fig.  368). 

(1)  Use  the  theorem  of  Pythagoras. 

(2)  Use  the  tangent  function. 


REGULAR  POLYGONS  AND  THE  CIRCLE 


287 


11.  Express  in  terms  of  the  radius  r  the  perimeters — 
(a)  of  the  regular  inscribed  and  circumscribed  hexagon, 
(6)  of  the  equilateral  inscribed  and  circumscribed  triangles. 

Show  that  the  perimeters  vary  directly  as  the  radii. 

443.  Problem:  To  inscribe  a  regular  decagon  in  a  given 
circle. 

Analysis:  Into  how  many  equal  arcs  must  the  circle 
be  divided? 

How  large  are  the  central  angles  intercepting  these 
arcs? 

Construction:  The  construction 
of  an  angle  of  36°  depends  upon 
the  problem  of  dividing  a  segment 
into  mean  and  extreme  ratio. 
(See  §  315,  exercise  8.) 

Draw  the  radius  AO,  Fig.  369. 

Divide  AO  into  mean  and  ex- 
treme ratio  at  B,  making 

OA^OB^ 
OB    BA' 

With  A  as  center  and  radius  OB  draw  an  arc  at  C. 
With  the  same  radius  and  center  C  draw  an  arc  at  D. 
Similarly,  draw  arcs  at  E,  F,  G,  H,  I,  J,  and  K. 
Draw  AC,  CD,  etc. 
Polygon  ACD K  is  the  required  polygon. 

Proof:  Draw  BC  and  OC. 
OA    OB 


OA^AC 
AC    BA' 


Why 


288 


SECOND-YEAR  MATHEMATICS 


I.e.,  in  ABC  A  and  AOC  two  sides  of  one  are  propor- 
tional to  two  sides  of  the  other. 

Show  that  the  included  angle  A 
is  the  same  in  both  triangles. 

.-.       ABCAc^AAOC.    Why? 
EC    CA  ,^,     „ 

.-.     OC  'BC  =  OA  -CA.  Why? 

BC  =  CA.  Why? 

BC  =  OB.  Why? 

Denoting    A  AOC    by  x,   show  Fig.  369 

that  OCB  =  x  and  that  ABCA=x. 

Since   AOCA=/.OAC,  it  follows  that   Z.0AC  =  2x, 
.-.  2a;+2a;+x  =  180°.  Why? 

a:  =  36°. 

Show    that    polygon  ACD K    is    a    regular 

decagon. 

EXERCISES 

1.  To  circumscribe  a  regular  decagon  about  a  circle. 

2.  Show  how  to  inscribe  and  circumscribe  a  regular  pentagon 
in  a  given  circle. 

3.  To  inscribe  and  circumscribe  regular  polygons  having 
20,  40,  etc.,  sides. 

4.  Show  that  the  number  of  sides  of  the  polygons  in  exer- 
cise 3  may  be  expressed  by  the  formula  5  •  2^,  w  being  a  positive 
integer  or  zero. 

5.  Express  the  relation  between  the 
side  of  the  inscribed  decagon  and  the 
radius  of  the  circle. 

Denoting  AC  =  OB  by  a,  Fig.  370,  OA  by 
r,  then  BA=r—a. 

Show  that     -  = . 

a    r—a 

a^  =  r^—ra. 

.*.    a2+ra-r2=0. 


REGULAR  POLYGONS  AND  THE  CIRCLE 


289 


Solving  by  means  of  the  quadratic  formula, 

„ — 

„     -r±rl/5 


=  2(- 


1=^/5) 


Show   that   the   minus   sign  before  the 
radical  cannot  be  used  in  this  'problem. 

:,    a=|('v/5-l)=^('l.236)  =  .618r. 

J6.  Show  that  the  side  of  a  regular 
inscribed  pentagon  is  equal  to^^lO— 2^/5 

Let  KC,   Fig.  371,  be  the  side  of  the 

pentagon,  KA  and  AC  sides  of  the  decagon. 

Denote  KF  by  6,  OK  by  r,  and  KA  by  a. 

Then, 


or 


r2-62andOF  =  Vr2-62. 


h\ 


Since 


KF'  =  KA'-FA', 

h''=a?-{r-V^i^^^y.        Why? 


Substituting  for  a^  its  equal,   ^(v^5  — l)j      (exercise  5)  and 
solving  for  h,  we  have — 


6  =  -l/ 10-21/5 


2&  =  ^v'l0-2i/5. 


J7.  Show   that    an  approximate   value   of   1^   10— 2V5  is 
2.351+. 

8.  Using  the  sine  function,  find  the  side  of  the  regular  in- 
scribed pentagon;  decagon. 

Notice  the  advantage  of  the  trigonometric  method  over  the 
algebraic  methods  used  in  exercises  5  and  6. 

9.  A  man  has  a  round  table  top  which  he  wishes  to  change 
into  the  form  of  a  pentagon  as  large  as  possible.  The  diameter 
of  the  top  is  2  J  feet.    What  is  the  length  of  the  cut  required  ? 


290  SECOND-YEAR  MATHEMATICS 

444.  Problem:  To  construct  a  regular  15-side  in  a 
given  circle. 

Analysis:  The  circle  must  be  divided  into  15  equal 
arcs.  How  large  are  the  central  angles  intercepting  these 
arcs? 

Notice  that  24°  =  60° -36°. 

This  suggests  the  following  con- 
struction : 


Construction:   At  0  on  OA  con- 
struct an  angle  of  60°,  Fig.  372. 

At  0  on  OA  construct  an  angle  of 
36°,  as   ZAOC.  Fig.  372 

Then  ZC05  =  24°. 

.'.  CB  may  be  taken  as  the  side  of  the  regular  in- 
scribed 15-side. 

EXERCISES 

1.  Show  how  to  construct  regular  inscribed  and  circum- 
scribed polygons  having  30,  60,  120  ....  sides. 

t2.  Show  that  the  number  of  sides  of  the  polygons  in  exer- 
cise 1  is  given  by  the  formula  15-2"  where  n  is  a  positive  integer, 
or  zero.f 

t  Gauss  (1777-1855),  a  German  mathematician,  proved  that 
by  the  use  of  an  unmarked  straight  edge  and  a  compass  a  circle  can 
be  divided  into  (2^+1)  equal  parts,  k  being  a  number  that  makes 
2^+1  a  prime  number. 

Denoting  2^-1-1  by  n,  we  have 

For  A;  =  1,  n  =  3,  a  prime  number. 

For  k==2,  n  =  5,  a  prime  number. 

For  A;  =3,  n  =  9,  not  a  prime  number. 

For  fc  =  4,  71  =  17,  a  prime  number. 

For  fc  =  5,  n  =  33,  not  a  prime  number,  etc. 


11^ 


CAEL  PEIEDRICH  GAUSS 


CARL   FRIEDRICH    GAUSS 


CARL  FRIEDRICH  GAUSS  was  born  at  Brunswick, 
Germany,  April  30,  1777,  and  died  at  Gottingen, 
February  23,  1855.  His  father  was  a  bricklayer  and  did 
not  sympathize  with  the  son's  aspirations  for  an  edu- 
cation. Coupled  with  this  was  the  fact  that  the  schools  of 
Gauss's  day  were  very  poor;  but  in  spite  of  parental  disap- 
proval and  very  inadequate  schools  he  became  one  of  the 
greatest  mathematicians  of  all  time. 

Gauss  had  a  marvelous  aptitude  for  calculation,  and  in  later 
years  used  to  say,  perhaps  only  as  a  joke,  that  he  could  reckon 
before  he  could  talk.  He  owed  his  education  to  the  fact  that 
one  of  his  teachers,  named  Bartels,  drew  the  attention  of  the 
reigning  duke  of  Brunswick  to  the  remarkable  talents  of  the 
boy.  The  duke  provided  for  him  the  means  of  obtaining  a 
liberal  education.  As  a  boy  Gauss  studied  the  languages  with 
quite  as  much  success  as  mathematics. 

When  only  nineteen.  Gauss  discovered  a  method  of  inscrib- 
ing a  regular  polygon  of  seventeen  sides  in  a  circle.  This 
encouraged  him  to  pursue  mathematical  studies.  He  studied 
at  Gottingen  from  1795  to  1798.  He  made  many  of  his  most 
important  discoveries  while  yet  a  student.  His  favorite  study 
was  higher  arithmetic.  In  1798  he  went  back  to  his  home 
town  of  Brunswick,  and  for  a  few  years  earned  a  scanty  living 
by  private  tuition. 

In  1799  Gauss  published  a  demonstration  of  the  important 
theorem  that  every  algebraical  equation  has  a  root  of  the  form 
a-\-bi,  and  in  1801,  a  volume  on  higher  arithmetic.  His  next 
great  performance  was  in  the  field  of  astronomy.  He  invented 
a  method  for  calculating  the  elements  of  a  planetary  orbit  from 
three  observations,  by  so  powerful  an  analysis  of  existing  data 
as  to  place  him  in  the  first  rank  of  theoretical  astronomers. 

In  1807  he  was  appointed  professor  of  mathematics  and 
director  of  the  observatory  at  Gottingen.  He  retained  these 
offices  until  his  death.  ^  He  was  devoted  to  his  work.  He 
never  slept  away  from  his  observatory  except  on  one  occasion 
when  he  attended  a  scientific  congress  in  Berlin.  As  a  teacher 
he  was  clear  and  simple  in  exposition,  and  for  fear  his  auditors 
might  not  get  his  train  of  thought  perfectly  he  never  allowed 
them  to  take  notes.  His  writings  are  more  difficult  to  follow, 
for  he  omitted  the  developmental  details  that  he  was  so  careful 
to  supply^  in  his  lectures.  His  memoirs  in  astronomy,  in 
geodesy,  in  electricity  and  magnetism,  in  electrodynamics, 
and  in  the  theories  of  numbers  and  celestial  mechanics  are 
all  epoch-making.  Most  of  the  whole  science  of  mathematics 
has  undergone  a  complete  change  of  form  by  virtue  of  Gauss's 
work. 

Gauss  was  the  first  to  develop  a  real  mathematical  theory 
of  errors.  He  introduced  the  geometrical  theory  of  complex 
numbers  into  Germany.  He  was  the  first  to  use  the  term 
"complex  number"  in  the  sense  it  has  today.  _  He  used  the 
symbol  =to  signify  congruence.  A  good  description  of  Gauss's 
important  work  on  the  inscription  of  a  regular  polygon  in  a 
circle  may  be  read  in  §  35  of  Miller's  Historical  Introduction 
to  Mathematical  Literatxire  (Macmillan). 

The  last-mentioned  work,  pp.  241-43,  and  also  both  Ball's 
and  Cajori's  Histories,  give  brief  accounts  of  Gauss  and 
bis  work. 


REGULAR  POLYGONS  AND  THE  CIRCLE 


291 


|3.  The  following  is  a  practical  method  of  constructing  the 
side  of  a  regular  10-side  and  5-side. 


Construction:  Draw  the  diameter  AB, 
Fig.  373. 

Draw  OC±AB. 

Bisect  OB  at  D.  ^ 

With  center  at  D  and  radius  DC  draw 

the  arc  CE. 

Draw  the  straight  line  CE. 

The  sides  of  triangle  EOC  are  equal 
to  the  sides  of  a  regular  hexagon,  penta- 
gon, and  decagon,  respectively. 


Fig.  373 


Proof:  I.  CO  =  r  and  is  equal  to  the  side  of  the  regular  inscribed 
n. 

II.  CW  =  r^+^. 


CD=lVl, 


EO 


:^D-OD  =  CD-OD=|l/5-^  =  ^(l/5-l)  . 


Hence,   EO  is  the  side  of  the  decagon. 
(See  §  443,  exercise  5.) 

r2 


III 


.  ^C'  =  r2+^eO'  =  r2+^  (6-2 1/5) 


4r2+6r2-2rV5 


EC  =  W\{)-2Vb,  the  side  of  the  pentagon. 

(See  §  443,  exercise  6.) 

4.  Find  the  side  of  a  decagon  inscribed  in  a  circle  of  radius 
8;  10;   15;  a. 

t5.  The  side  of  an  inscribed  pentagon  is  18.8  inches.    Find 
the  radius  of  the  circumscribed  circle. 


6.  The  side  of  an  inscribed  decagon  is  14 .  83  inches.     Find 
the  radius  of  the  circumscribed  circle. 


292 


SECOND-YEAR  MATHEMATICS 


J7.  //  at  the  midpoint  of  the  arcs  subtended  by  the  sides  of  a 
given  regular  inscribed  polygon,  tangents  are  drawn  to  the  circle, 
they  are  parallel  to  the  sides  of  the  given  polygon  and  form  a  regular 
circumscribed  polygon. 

To  prove  that  AB  \\  A'B',  Fig.  374,  draw  the  radius  OP'  to  the 
contact  point  of  A'B\    Show  that  AB 
and  A'B'  are  both  perpendicular  to  OP'. 

To  prove  that  A'B'C'D'E'  is  regu- 
lar, show  that  fQ'  =  Q^'  =  RS',  etc. 

8.  In  Fig.  374,  prove  that  points 
0,  B,  and  B'  are  on  a  straight  line.    ^ 

Prove  that  B  and  B'  lie  on  the 
bisector  of  ZP'OQ'. 

9.  Express  8  as  a  theorem. 

445.  Theorem:  A  circle  may  he  circumscribed  about 
any  given  regular  polygon. 

Given  the  regular  polygon 
ABCD ,  Fig.  375. 

To  construct  a  circle  circum- 
scribed about  ABCD 

Construction:  Construct  a  circle 
through  A,  B,  and  C. 

This  is  the  required  circle. 

Proof:  It  is  to  be  proved  that 

the  circle  ABC  passes  through  D,  E, 

x-\-y  =  z-\-u.       Why? 

y  —  z.  Why  ? 

.*.     x  =  u.  Why? 

Prove       AAOB^ACOD. 

.'.     A0  =  OD  and  the  circle  passes  through  D. 
Similarly,  it  may  be  shown  that  the  circle  passes 
through  E,  F,  etc. 


etc. 


REGULAR  POLYGONS  AND  THE  CIRCLE         293 

446.  Theorem:  A  circle  may  be  inscribed  in  any  given 
regular  polygon. 

Given  the  regular  polygon 
ABC ,  Fig.  376. 

Requured  to  inscribe  a  circle 
within  ABC 

Construction:  Construct  the 
center,  0,  of  the  circumscribed 
circle. 

Draw  OK±AB. 

With  0  as  center  and  radius  OK  draw  circle  KLM 

This  is  the  required  circle. 

Proof:  Draw  the  circumscribed  circle  ABC. 

Draw  OP±AE. 

Since  chord  AS  =  chord  AE,  it  follows  that  OK  =  OP. 
Why? 

Hence,  circle  KLM  passes  through  P.        Why  ? 

.•.     AE  is  tangent  to  the  circle.         Why  ? 

Similarly,  show  that  ED,  DC,  etc.,  are  tangents  to 
circle  KLM 

447.  Theorem:  The  perimeter  of  a  regular  inscribed 
2n-side  is  greater  than  the  perimeter  of  the  regular  n-side 
inscribed  in  the  same  circle.     Prove. 

448.  Theorem:  The  perimeter  of  a  regular  circum- 
scribed 2n-side  is  less  than  the  perimeter  of  the  regular 
n-side  circumscribed  about  the  same  circle.     Prove. 

449.  Two  important  facts  follow  from  the  theorems 
in  §§  447  and  448,  viz.: 

1.  The  perimeter  of  the  regular  inscribed  polygon 
increases  as  the  number  of  sides  increases. 


294  SECOND-YEAR  MATHEMATICS 

2.  The  perimeter  of  the  regular  circumscribed  polygon 
decreases  as  the  number  of  sides  increases. 

The  Length  of  the  Circle 

450.  In  the  following  discussion  it  will  be  shown  that 
by  increasing  the  number  of  sides  of  regular  inscribed 
and  circumscribed  polygons,  the  perimeters  approach 
each  other  more  and  more,  and  that  the  decimal  fractions 
expressing  these  two  perimeters  can  be  made  to  agree 
to  a  greater  and  greater  number  of  decimal  places. 

It  is  easily  proved  that  the  length  of  a  circle  is  greater 
than  the  perimeter  of  any  inscribed  polygon.  We  will 
assume  that  the  length  of  a  circle  is  less  than  the  perim- 
eter of  any  circumscribed  polygon. 

Hence,  the  length  of  a  circle  lies  between  the  lengths 
of  the  perimeters  of  any  pair  of  inscribed  and  circumscribed 
polygons. 

The  determination  of  the  length  of  the  circle  is 
obtained   very   simply   by  means  of  ^..^— --^ 

trigonometry :  y^  ^s. 

Let  AB,  Fig.  377,  be  the  side  of  a     /  \ 

regular  inscribed  n-side.  [  o  \ 

Draw  OD±AB.  \        /fV       / 

e^  FiQ.  377 

(360\°      2 
^—j  -—  ,  a  denoting  the  side  of  the 

polygon  and  r  the  radius  of  the  circumscribed  circle. 

Hence,  a  =  sin  ( -^ )  ^.  Why  ? 

sin(— j    \d.     Why?..  (A) 


REGULAR  POLYGONS  AND  THE  CIRCLE         295 


From  Fig.  378  show  that  the  perimeter  P  of  the  circum- 
scribed polygon  is  given  by 


{-(I?)-] 


By  means  of  formulas  (A)  and  (B) 
the  perimeters  of  inscribed  and  circum- 
scribed polygons  may  be  computed, 
leading  to  the  determination  of 
approximate  values  of  the  length  of  ^^^  g^g 

the  circle. 

Make  the  computations  and  compare  your  results 
with  the  results  given  in  the  following  table: 


Number 
of  sides 

Perimeter  of 
inscribed  polygon 

Length  of  circle,  I 

Perimeter 

of  circumscribed 

polygon 

3 

2.5980... 

•  d 

2.5d<l<5.2d 

5.1963....  'd 

4 

2.8284... 

'd 

2.Sd<l<4:.0d 

4.0000....  'd 

5 

2.9390... 

'd 

2.9d<l<S.7d 

3.6325....  'd 

6 

3.0000... 

'd 

S.0d<l<S.5d 

3.4644....  -d 

7 

3.0359... 

'd 

S.0d<l<3Ad 

3.3691....  'd 

8 

3.0614... 

'd 

S.Od<l<SAd 

3.3137....  -d 

12 

3.1058... 

'd 

S.ldKKS.Sd 

3.2153....  'd 

18 

3.1248... 

.  'd 

l  =  S.ld,  approxi- 
mately 

3.1734....  'd 

90 

3.1410....  'd 

l  =  3.14:ld,  approxi- 
mately 

3.141 -d 

The  table  above  shows  how  the  decimal  fractions 
expressing  the  perimeters  agree  more  and  more  closely 
as  the  number  of  sides  of  the  polygon  is  increased. 


296 


SECOND-YEAR  MATHEMATICS 


The  following  table,  which  gives  the  decimal  fractions 
to  six  places,  shows  the  approach  of  the  perimeters  still 
better : 


Pi* 

=  4.000000 

■d 

Pe 

=  3.464121.. 

•d 

Ps 

=  3.313708.. 

'd 

Pl2 

=  3.215390.. 

■d 

Pl6 

=  3.182598.. 

■d 

P24 

=  3.159659.. 

•d 

P32 

=  3.151725.. 

•d 

P48 

=  3.146086.. 

■d 

P64 

=  3.144118.. 

■d 

P96 

=  3.142714.. 

'  d 

Pl28 

=  3.142224.. 

•d 

Pl92 

=  3.141873.. 

•  d 

P256 

=  3.141750.. 

■d 

P348 

=  3.141662.. 

■  d 

P4'=2 

P6    =3 
P8    =3 
Vn  =3. 
P16  =3. 
P24  =3. 

P32  =3. 
p48  =3. 
P64  =3. 
P96  =3. 
?>128=3. 
Pl92  =  3  . 
P256  =  3  . 
P348  =  3  , 


.828427. 
.000000 
.061467. 
.105828. 
.121445. 
.132623. 
.136548. 
.139350. 
. 140331 . 
.141032. 
.141277. 

141452. 

141514. 

141557. 


The  last  two  perimeters  agree  to  three  decimal  places'. 
Thus,  the  length  of  the  circle  of  diameter  d  which  lies 
between  these  perimeters  is  found  correct  to  three  decimal 
places.     It  equals  3 .  141 ...  •  d. 

As  the  perimeters  of  the  inscribed  and  circumscribed 
polygons  with  increasing  numbers  of  sides,  approach 
each  other  in  length,  both  of  them  approach  more  and 
more  closely  the  length  of  the  circle.  But  however  close 
the  length  of  the  perimeter  of  any  polygon  may  come 
to  the  length  of  the  circle,  there  is  always  another  polygon 
the  perimeter  of  which  comes  still  closer  to  the  length 
of  the  circle;  and  for  every  number  given  as  expressing 
the  difference  between  any  perimeter  and  the  circle  we 
can  find  a  polygon  whose  perimeter  differs  from  the  circle 
by  less  than  that  number.  This  is  expressed  by  saying 
that  the  perimeters  of  the  inscribed  and  circumscribed 
polygons  approach  the  circle  as  a  limit. 


*  The  subscripts  indicate  the  number  of  sides  of  the  polygons. 


.REGULAR  POLYGONS  AND  THE  CIRCLE         297 

As  is  seen  by  the  table  on  p.  296,  the  value  of  this  limit 
can  be  expressed  more  and  more  closely  by  taking  poly- 
gons of  a  greater  and  greater  number  of  sides.  It  cannot, 
however,  be  determined  exactly. 

Continuing  to  increase  the  number  of  sides,  we  find 
in  the  table  above 

P8i92  =  3. 1415928..  'd 
and  2>8i92  =  3. 1415926..  -  d. 

From  this  it  is  seen  that  the  circle,  being  between 
Psm  and  p8i92,  can  be  expressed  by  0  =  3.141592..  •  d, 
approximately,  with  an  error  less  than  1  millionth. 

The  length  of  the  circle  is  therefore  a  multiple  of  the 
diameter,  which,  however,  may  not  be  exactly  expressed 

in  figures.     The   number  3.141592 by   which    d   is 

multiplied,  is  commonly  denoted  by  -n-  (the  first  letter  of 
7re/3t<^e/3€ta,  meaning  periphery  or  circumference). 

Thus,  C  =  ir(/, 

and  C  =  2irr 

are  the  formulas  expressing  the  length  of  the  circle  in 
terms  of  the  diameter  and  radius,  respectively.  For  our 
purposes  it  is  sufficient  to  use  7r  =  3 .  14,  or  7r  =  -Y-,  which  is 
equal  to  3 .  14  when  carried  out  to  two  decimal  places. 

451.  Historical  note.  The  determination  of  the  value  and 
of  the  nature  of  the  number  v  is  one  of  the  famous  problems  of 
geometry. 

16^2 


Ahmes  took 


-(?)■ 


Archimedes  (212-287  B.C.)  found  the  value  of  tt  to  be  such 
that  3;^  <7r  <3=^  by  finding  the  values  of  Pge  a.ndp%. 

Ptolemy  (150  a.d.)  calculated  ^  =  3+|T+Jr„  =  3. 14166. 

DU      OU 


298  SECOND-YEAR  MATHEMATICS 

At  the  end  of  the  sixteenth  century  Vieta  (1579  a.d.)  found 
the  value  of  tt  to  10  decimal  places,  and  Ludolph  van  Ceulen 
(1540-1610)  to  20,  32,  and  35  places.  The  value  of  tt  has  since 
been  carried  out  to  more  than  700  decimal  places,  to  30  places 
it  is  as  follows: 

3.141592653589793238462643383279+  (see  the  article 
*' Circle"  in  the  Encyclopaedia  Brittanica,  11th  ed.). 

It  was  shown  by  Lambert  (1728-1777)  that  the  number  tt 
cannot  be  expressed  exactly  in  terms  of  integers  and  hence  is 
not  a  rational  number. 

Lindemann  (1882)  proved  that  tt  belongs  to  a  class  of  num- 
bers called  transcendental,  numbers  which  do  not  satisfy  any 
algebraic  equation  with  rational  coefficients. 

EXERCISES 

1.  The  length  of  a  circle  is  100  inches.     Find  the  radius. 

2.  Show  that  the  lengths  of  two  circles  are  to  each  other  as  the 
radii  or  as  the  diameters. 

3.  The  distance  around  one  of  the  famous  large  trees  in 
California  is  about  100  feet.     Find  the  diameter. 

4.  The  radius  of  a  fly  wheel  of  an  engine  is  9  feet.  If  the 
wheel  makes  40  revolutions  per  minute,  what  is  the  rate,  in 
feet,  per  minute  of  a  point  on  its  outer  rim  ? 

5.  The  size  of  a  man's  hat  is  indicated  by  the  number  of 
inches  in  the  diameter  of  a  circle  of  length  equal  to  the  distance 
measured  around  the  head  where  his  hat  rests.  What  size  of  hat 
does  a  man  need,  the  distance  around  whose  head  is  22f 
inches  ? 

6.  Measure  the  distance  around  your  own  head  and  calculate 
the  size  of  hat  you  need. 

7.  A  trick  circus  rider  performed  on  a  tall  bicycle  one  turn  of 
whose  driving  wheel  carried  the  bicycle  62 . 8  ft.  forward.  How 
tall  was  the  wheel  ? 


REGULAR  POLYGONS  AND  THE  CIRCLE        299 

8.  A  circular  pond  is  2640 . 1  yd.  in  circumference.  Find 
the  diameter. 

452.  Historical  note.  Regular  polygons  have  been  used  for 
decorative  purposes  since  the  beginning  of  mathematical  history. 
Only  such  regular  polygons  as  result  from  the  division  of  the 
circle  into  4  equal  parts,  i.e.,  squares,  octagons,  etc.,  were  known 
in  Egypt  before  the  Eighteenth  Dynasty.  About  this  time  the 
dodecagon  appeared  on  presents  sent  to  Pharaoh  by  his  Asiatic 
subjects.  Since  the  Nineteenth  Dynasty  chariot  wheels  with 
six  spokes  are  sho"WTi  on  mural  reliefs,  and  very  rarely  with  four 
or  eight  spokes.  The  knowledge  of  the  sextuple  division  of  the 
circle  was  brought  to  Egypt  from  Babylon,  though  it  is  not 
known  at  what  date  this  occurred.  The  Chaldaeans  had  a  strong 
bias  in  favor  of  six  and  its  multiples. 

The  Greeks  advanced  the  knowledge  of  regular  polygons. 
The  Pjrthagoreans  thoroughly  reworked  Egyptian  and  Baby- 
lonian knowledge  and  extended  it  by  original  research.  They 
taught  how  to  calculate  the  central  angle  for  all  n-gons.  It  is 
not  definitely  known  whether  the  Pythagoreans  could  construct 
the  regular  pentagon,  though  they  used  the  pentagram  (the  star 
pentagon)  as  a  symbol  of  secrecy,  and  at  least  studied  the  penta- 
gon. At  all  events  the  Greek  mathematicians  by  the  time  of 
Eudoxus  (408-355  B.C.)  were  masters  of  the  division  of  a  line 
into  mean  and  extreme  ratio,  upon  which  the  construction  of  the 
regular  decagon  depends. 

That  the  side  of  a  regular  inscribed  hexagon  is  equal  to  the 
radius  of  the  circle  was  known  in  substance  to  the  ancient  Baby- 
lonians. Hippocrates  (440  b.c.)  mentions  this  property  of 
the  hexagon  as  a  well-known  theorem.  The  mode  of  calculating 
the  sides  of  our  most  familiar  regular  polygons  was  known 
by  the  time  of  Hero  of  Alexandria  (first  century,  B.C.). 

Antipho  (430  B.C.)  was  the  first  to  make  use  of  the  regular 
inscribed  polygon  to  approximate  the  area  and  length  of  the 
circle.  Bryso,  a  contemporary  of  Antipho,  improved  on  the 
latter's  method,  and  the  theory  was  very  greatly  extended  by 
Archimedes.    The  latter  had  a  method  of  calculating  the  side 


300  SECOND-YEAR  MATHEMATICS 

of  a  2?i-gon  from  the  side  of  an  n-gon.    By  means  of  regular 
polygons  he  shut  tt  in  between  the  limits  of  3y  and  Syy. 

After  Archimedes  no  further  advance  in  the  theory  of 
regular  polygons  was  made  nntil  the  thirteenth  century.  Jor- 
danus  Nemorarius  (1237  a.d.)  did  not  seek  to  square  the  circle 
by  the  aid  of  regular  polygons,  as  most  later  writers  had  done, 
but  rather  to  derive  relations  between  the  perimeters  and  areas 
of  regular  inscribed  and  circumscribed  polygons  of  n  and  2n 
sides. 

Summary 

453.  The  chapter  has  taught  the  meaning  of  the  fol- 
lowing terms: 

regular  polygon,  circumscribed  polygon,  inscribed  polygon 

454.  The  following  theorems  may  be  used  to  prove 
that  inscribed  or  circumscribed  polygons  are  regular: 

1.  If  a  circle  is  divided  into  equal  arcs  the  chords  sub- 
tending these  arcs  form  a  regular  inscribed  polygon. 

2.  If  the  midpoints  of  the  arcs  subtended  by  the  sides 
of  a  regular  inscribed  polygon  of  n  sides  are  joined  to  the 
adjacent  vertices  of  the  polygon,  a  regular  inscribed  polygon 
of  2n  sides  is  formed. 

Z.  If  a  circle  is  divided  into  equal  arcs,  the  tangents 
drawn  at  the  points  of  division  form  a  regular  circumscribed 
polygon. 

4.  If  tangents  are  drawn  to  a  circle  at  the  midpoints 
of  the  arcs  terminated  by  consecutive  points  of  contact  of 
the  sides  of  a  regular  circumscribed  polygon,  a  regular  cir- 
cumscribed polygon  is  formed  having  double  the  number  of 


455.  Other  theorems  proved  in  the  chapter  are: 
1.  If  at  the  midpoints  of  the  arcs  subtended  by  the  sides 
of  a  given  regular  inscribed  polygon,  tangents  are  drawn  to 


REGULAR  POLYGONS  AND  THE  CIRCLE         301 

the  circle,  they  are  parallel  to  the  sides  of  the  given  polygon 
and  form  a  regular  circumscribed  polygon. 

2.  A  circle  may  he  circumscribed  about  any  given  regular 
polygon. 

3.  A  circle  may  be  inscribed  in  any  given  regular  polygon. 

4.  The  perimeter  of  a  regular  inscribed  2n-side  is  greater 
than  the  perimeter  of  the  regular  n-side  inscribed  in  the  same 
circle. 

5.  The  perimeter  of  a  regular  circumscribed  2n-side  is 
less  than  the  perimeter  of  the  regular  n-side  circumscribed 
about  the  same  circle. 

456.  The  chapter  has  taught  how  to  inscribe  m,  and  to 
circumscribe  about  a  circle  the  following  regular  polygons : 
square,  hexagon,  decagon,  15-side. 

Other  regular  inscribed  and  circumscribed  polygons 
may  be  obtained  by  dividing  the  arcs  of  the  circle  into 
two  or  more  equal  parts,  and  then  joining  the  points  of 
division  of  the  circle  successively  by  line-segments. 

457.  The  side  and  perimeter  of  a  regular  inscribed  or 
circumscribed  polygon  may  be  expressed  in  terms  of  the 
radius  of  the  circle.  The  side  and  perimeter  vary  directly 
as  the  radius. 

458.  The  length  of  a  circle  is  expressed  by  the  formula 

C  =  Trd,  or  C  =  2irr, 


CHAPTER  XVIII 

COMPARISON  OF  AREAS.    LITERAL  EQUATIONS. 
AREA  OF  THE  TRIANGLE.    FACTORING 

Comparison  of  Areas 
459.  Theorem:     Parallelograms    having    equal    bases 
and  equal  altitudes  are  equal* 

D  CD'         d"       C  c" 

V 7 


7 — 
h 


Fig.  379 

Given  parallelograms  ABCD  and  A'B'C'D\  Fig.  379, 
having  equal  altitudes  h,  and  equal  bases  6. 
To  prove  that  ABCD=^A'B'C'D'. 

Proof:  Imagine  ABCD  placed  upon  A'B'C'D',  so 
that  AB  coincides  with  A'B',    Why  can  this  be  done? 

Then  DC  must  fall  in  the  same  hne  as  D'C,  for  the 
parallelograms  have  equal  altitudes. 

Prove  that     AD'D"A'^C'C"B'  (s.a.s.). 

But  D'A'B'C'-^D'A'B'C', 


.      A'B'C''D''  =  A'B'C'D' 
(equals  subtracted  from  equals  give  equals). 

ABCD  =  A'B' CD'.        Why?       ' 

460.  Theorem:  A  'parallelogram  is  equal  to  a  rectangle 
having  the  same  base  and  altitude. 
Apply  the  theorem  in  §  459. 

*  Equal  is  here  used  in  the  sense  of  equal  in  area,  or  equivalent. 
302 


AREAS.     LITERAL  EQUATIONS.     FACTORING      303 


461.  Theorem:    A    triangle 
parallelogram  having  the  same  base 
and  altitude. 

Use  the  theorem  that  a  diagonal 
divides  a  parallelogram  into  congru- 
ent triangles  (Fig.  380). 


equal    to    one-half   a 


Fig.  380 


462.  Theorem  of  Pjrthagoras:  The  square  on  the 
hypotenuse  of  a  right  triangle  is  equal  to  the  sum  of  the 
squares  on  the  sides,  including  the  right  angle. 

Let  ABC,  Fig.  381,  be  a 
right  triangle  having  a  right 
angle  at  C.  Let  Si,  S2,  and  S 
denote  the  squares  on  the  sides 
a,  h,  and  c,  respectively. 

To  prove  8  =  81+82. 

Proof:     Draw    CD±AB, 

dividing  8  into  rectangles  i^i 
and  R2. 

Draw  AE  and  CF. 

Show  that  triangle  EBA 
and  square  aSi  have  equal  bases 
and  altitudes. 

Then  triangle  EBA  =  ^81. 

Similarly,  prove  that 
triangle      FBC  =  \Ri. 

But  AABE^AFBC. 

For  EB  =  BC.         Why? 

AB  =  BF.         Why? 

And  Z  ABE  =  Z  FBC.  Why  ? 

From  (1),  (2),  (3)  we  have  i8i  =  ^Ri. 

Therefore  *Si  =  7^i. 

Similarly,  draw  BG  and  CH,  and  prove  82  =  Rz. 

Therefore  S^Si+S^. 


Why? 


(1) 

(2) 
(3) 


(4) 


304  SECOND-YEAR  MATHEMATICS 

463.  Theorem:    The  sum  of  the  squares  of  two  sides 
of  a  triangle  is  equal  to  twice  the  square  of  one-half  of  the 
third     side    increased    hy    twice 
the  square  of  the  median  to  the  ^^^ 

third  side*  ^^  /  I  \a 

Given    AABC    having    the       ^{:^..<^.^iT'i.^.\^ 
median  m  to  the  side  c,  Fig.  382.  ^  ^ 

/^\  2  Fig.  382 

To  prove  that  a^-f  62  =  2(|j  -{-27n\ 

Proof:  a2  =  {^ -^m^-  2{^m'.       §  240. 

h'  =  (^\m'+2{^m'.        §241. 

.-.    a2+62  =  2(0'+2^?^2.        Why? 

Exercises  in  Literal  Equations  in  One  Unknown 

1.  Show  that  the  length  of  the  median  to  a  side  of  a  triangle 
may  be  expressed  in  terms  of  the  sides  of  the  triangle  by 
means  of  the  following  formula: 


|^2  +  52_2Q  ______ 

mc=^' 2" ^  =  2^2a2+2fe2_c2 

2.  Find  nic  when  a,  b,  and  c  are  respectively, 

(1)  6,  10,  8  1(2)  5,  13,  12  J(3)  9,  15,  12 

3.  Express  Wo  in  terms  of  the  sides  a,  h,  and  c  of  the 
triangle  ABC. 

*  This  theorem  was  added  to  elementary  geometry  by  Pappus 
who  lived  and  taught  at  Alexandria  at  the  end  of  the  third  century 
A.D.  It  enables  us  to  find  the  medians  of  a  triangle  when  the  lengths 
of  the  sides  are  known. 


AREAS.     LITERAL  EQUATIONS.     FACTORING      305 

Solve  the  following  equations  for  x  or  y: 

4.  ax—bx  =  a  —  b 
Combining  the  terms  in  x,  {a  —  b)x  =  a  —  b. 
Dividing    both   sides    by  a  —  b,x  =  l. 

6.  x^-c^  =  3c^-2cx-\-x^ 

6.  x^—2ax+c  =  a-\-3ax-\-x^ 

7.  a^-\-ax-{-cx  —  ac  =  0 

8.  -  3rx + 2cm  = —3cx-\-  2rm 

9.  -¥x+a'^x=-a+b 

10.  cx4-ax  =  a2+c2+2ac 

11.  Sx-ax  =  a^+9-U 

12.  s^x-{-r^x—2rsx  =  r'^  —  s'^ 

13.  (m+n)x+{m—n)x  =  2?n^ 

14.  -9(2/-a)+6(22/+a)  =  -2(y+a) 
tl5.  m{my+n)-\-n'^  =  n{my-^n)-{-m'^ 

..„   3;  ,  o      1  «^    2x-b    x  —  2s 

17.  -4-3  =  ^  24.  ^— 7-  =  ^— TT 
o          2a  6x+s     3x+6 

.^    X  ,  X      1  .^^    2x— a    X  — a     a 

18.  -+7  =  -T  t25.  ■ x-  =  - 

a    0    ah  c  a       c 

^-        X  ,  x     — 62+a2  „^    ca;— d  ,  rfx— c    o^c—d 

a    b         ab  ax          ex  cdx 

+on    a;  ,-x_6a-86  x„-       2a;+a     -x-a_     a 

t2i.  ?-x=i+l-2  128.  ?!::^+^±^=2^_^ 

a  a^    a  ex  c        c     x 

-a+b     -a-b^  29.  ^^  =  -!L. 

X  X  x—m    x—n 

23.  ?-x=-c+-l  t30.  5±!!?  =  !?^ 

c  c  x—n     m— 71 


306 


SECOND-YEAR  MATHEMATICS 


Systems  of  Linear  Equations  in  Two  Unknowns  Having 
Literal  Coefficients 


464.  Solve  for  x  and  y: 

'aH-\-h^]i^ 

a+b 

(1) 

.ahx  —  ahy  = 

-h  —  a 

(2) 

aX(l) 

aH-\-ab^y  =  a^-{-ab 

(3) 

6X(2) 

a¥x-ab^y  =  ¥-ab 

(4) 

(Add.  Ax.) 

_1 

(5) 

a^+ab^    a(a2-f62) 

a 

Find  the  value  of  y,  first  by  eliminating  the  x-terms  from 
equations  (1)  and  (2);  and  then  by  substituting  a;  =  -  in  equa- 
tion  (2).    Check  your  results. 


EXERCISES 

Solve  each  of  the  following;  then  check: 
^    {  x+y=l 
\ax—by=0 

'cx+ny=l 


3. 


6. 


ax  —  by  =  0 

(ax-\-by  =  h 
\bx-\-ay  =  k 

cx-\-dy  =  2cd 
bx—cy  =  d—c 

(  ax+by  =  2ab 


\2bx+Say  =  2¥+Sa^ 

^    (aiX-\-biy  =  Ci 
'  \a2X-{-b2y  =  C2 


•=7. 


t9. 


1  +  1.1 
X    y    n 


a  ,  b 
X    y 

X    y 


X     y 
X     y 


*  Equations  in  exercises  7,  8,  and  9  are  not  linear  in  x  and  y, 

but  in  -  and  - . 
X        y 


AREAS.     LITERAL  EQUATIONS.     FACTORING      307 

The  Area  of  the  Triangle 

465.  Since  all  plane  figures  formed  by  straight  Unes 
may  be  divided  into  triangles,  it  is  important  to  obtain 
formulas  for  computing  the  area  of  a  triangle  from  given 
parts.  All  other  figirres  may  then  be  measured  by  means 
of  the  triangle.  We  are  acquainted  with  the  following 
formula  which  gives  the  area  of  a  triangle  in  terms  of  the 
base  and  altitude: 

Theorem:  The  area  of  a  triangle  is  equal  to  one-half 
the  product  of  the  base  and  altitude.     (§  56). 

AABC=lb'h 

466.  The  area  of  a  triangle  may  be  expressed  in  terms 
of  two  sides  and  the  included  angle. 

For,  from  Fig.  383,  AABC  =  ih  •  h. 

Since  sinA=-, 

c^ 

it  follows  that  h  =  c  sin  A. 

By  substitution,       AABC  =  ^  be  sin  A, 

This  may  be  expressed  as  a 
theorem  as  follows: 

Theorem:  The  area  of  a  tri- 
angle is  equal  to  one-half  the  prod- 
uct of  two  sides  by  the  sine  of  the 
included  angle. 

EXERCISES 

Show  that  the  area  of  an  equilateral  triangle  of  side  a 

is  equal  to  -rVZ. 

Show  that  the  area  of  a  regular  hexagon  of  side  a  is 
3l/3  „ 


308 


SECOND-YEAR  MATHEMATICS 


467.  Triangles  inscribed  in,  or  circumscribed  about,  a 
circle  are  frequently  met. 

The  areas  of  such  triangles  may  be  expressed  in  terms 
of  the  sides  and  the  radius  of 
the  circle  as  follows : 

Let  0,  Fig.  384,  be  the 
center  of  the  inscribed  circle. 

Draw  OA,  OB,  and  OC, 
dividing  triangle  ABC  into 
three  triangles  whose  sum  is 
AABC. 

Show  that  ACOA  =  ^r  '  b 

AA05  =  Jr  .  c 
ABOC^^r-a 

.-.     AABC  =  lr  (a+b+c). 

Hence,  the  area  of  a  triangle  is  equal  to  the  product 
of  one-half  the  perimeter  by  the  radius  of  the  inscribed 
circle. 

It  is  customary  to  denote  J(a+6+c)  by  the  symbol  s. 

Then,  AABC  =  rs. 

468.  Theorem:    The  area  of  a  triangle  is  equal  to  the 
product  of  the  three  sides  divided  by 
four  times  the  radius  of  the  circum- 
scribed circle. 

For,  let  ABC,  Fig.  385,  be  an 
inscribed  triangle. 

Draw  the  diameter  BE. 
Join  EC. 

AABC^ib'h 
Show  ABDA  CO  ABCE 

h  _  c 
a~2r' 


Fig.  385 


Then, 


Why? 


AREAS.     LITERAL  EQUATIONS.     FACTORING      309 


,     ac 


ac 
2r' 


By  substitution,  AABC  =  ^  •  b 

4r 

EXERCISES 

1.  The  three  sides  of  a  triangle  are  14,  8,  and  12.  The 
diameter  of  the  circumscribed  circle  is  14.1.  Find  the  area  of 
the  triangle. 

2.  Denoting  the  area  of  a  triangle  by  T,  then  T = -r— .    Solve 

4r 

the  equation  for  r.  Find,  in  terms  of  T,  the  radius  of  the  circle 
circumscribed  about  a  triangle  whose  sides  are  17,  10,  and  9; 
8,  8,  and  8;   15,  20,  and  25. 

3.  Using  the  facts  that  the  area  of  a  triangle  is  ^bh  and  -^ , 

d  being  the  diameter  of  the  circumscribed  circle,  find  a  formula 
for  the  altitude  to  the  side  b  in  terms  of  the  other  sides  and  the 
diameter. 

4.  The  sides  of  a  triangle  are  12,  10,  and  8.  The  area  is 
39 . 7.    Find  the  diameter  of  the  circumscribed  circle. 

5.  The  angles  of  a  right  triangle  are  to  each  other  as  1:2:3 
and  the  altitude  on  the  hypoteniise  is  6  feet.     Find  the  area. 

Je.  Heron  (1st  cen.  B.C.)  expressed  the  altitude  and  area 
respectively,  of  an  equilateral  triangle  as  h=a(l—^-Q—3\), 

and  A  =  aHi-\-T\)' 

Calculate  the  errors  of  Heron's  expressions. 

469.  The  area  of  a  triangle  may  be  expressed  in  terms 
of  the  sides  alone,  thus: 


310  SECOND-YEAR  MATHEMATICS 

Theorem:  The  area  of  a  triangle,  in  terms  of  its  sides 
is  V^s{s—a){s  —  b){s—c). 

B 

Given    in   triangle  ABC,  /A 

Fig.  386,  the  sides  a,  h,  and  c.  cy^     \  \ 

To  prove  that  the   area   of  /^  |    \ 

ABC  is  equal  to  ^  ^<:„_A:.'',V-i-"--V 
Vs{s  —  a){s  —  b){s  —  c)'  FiQ.  386 

Proof:  AreaA5C  =  |6-/i  (1) 

This  gives  the  area  in  terms  of  one  feide  and  the  alti- 
tude h,  which  is  not  known.  Let  us  now  express  h  in 
terms  of  the  sides  and  then  substitute  for  h  in  equa- 
tion (1). 

h^^c-'-ih-ay.        Why?  (2) 

h^  =  a^-a'\  Why?  (3) 

We  must  next  eliminate  a',  which  is  not  one  of  the  three 

sides. 

By  comparison,    c^  —  (6  —  a'Y  =  a^  —  a'^.  (4) 

Therefore,     c''-a^-¥+2ba'  =  0.  ^  (5) 

Solving  for  a',  we  find         a'  = ^ ,  (6) 

Substituting  in  (3)  the  value  of  a'  found  in  (6),  we  get 

Equation  (7)  expresses  /i^  in  terms  of  the  sides  a,  b, 
and  c. 

We  could  now  substitute  the  value  of  h  in  equation 
(1)  and  have  a  formula  for  the  area  of  ABC  in  terms  of 
a,  b,  and  c.  But  in  order  to  get  a  more  symmetrical  result, 
the  value  of  h^  in  (7)  will  be  changed  inform  before  sub- 
stituting in  (1). 


AREAS.     LITERAL  EQUATIONS.     FACTORING       311 

The  right  side  of  equation  (7),  being  the  difference  of 
two  squares,  may  be  factored  thus : 

^^V-^—2^)V 2b—)' 

Carrying  out  the  indicated  addition  and  subtraction 
within  the  parentheses,  we  have 

^  2b  '  2b  ' 

^= 26 25 •      ^^^• 

"'2b  2b         '  ^^^' 

Or  ,,_(a+5-c)(a+6+c)     {c-\-a-b){c-a+b)     ,  . 

^"^  '^  ~  26  '  26  •  ^^^ 

Let  a+6+c=2s. 

Subtracting  from  both  sides  of  this  equation  first  2c, 
then  2a  and  then  26,  we  have 


a+6-c  =  2s-2c  =  2(s-c] 

b-\-c-a  =  2s-2a  =  2(s-a)}-  (9) 

c-f  a-6  =  2s-26  =  2(s-6)J 


01 

a)\ 


Substituting  (9)  in  (8), 


2(s-c)  '  2s  •  2is-b)  ■  2{s-a) 
^  ~  462 

4:8  '  {s—a){s—b)(s—c) 

Therefore, 


h  =  jys{s-a){s-b){s-c)^        Why?         (10) 


312 


SECOND-YEAR  MATHEMATICS 


Substituting  (10)  in  (1), 
1 


ABC  =  ^h  '  -i/sis-a){s-b){s-c). 
Therefore, 

ABC  =  Vs{s-a)is-b){s-c)  . 


(11)* 


EXERCISES 

1.  The  sides  of  a  triangle  are  3,  5,  and  6.     Find  the  area. 
Using  formula  (11)  of  §  469, 

thearea  =  i/7^(7-3)(7-5)(7-6)  =  v'7  -4  -2  •  1  =2i/l4,or7.482, 
approximately. 

2.  The  sides  of  a  triangle  are  34,  20,  and  18.     Find  the  area. 

3.  The  sides  of  a  triangle  are  10,  6,  and  8.     Find  the  area. 

|4.  The  sides  of  a  triangle  are  90,  80,  and  26.     Find  the  area. 

JS.  The  sides  of  a  triangle  are  70, 
58,  and  16.     Find  the  area. 

470.  Altitudes  of  a  triangle. 
Denoting    the   altitudes   of    the 
triangle  ABC  to  the  sides  a,  6,    ^ 
and  c  by  ha,  h,  and  he,  respec- 
tively, Fig.  387,  show  that 

2  / 

hb  =  rVs{s  —  a){s  —  h){s  —  c) 


Fia.  387 


(See  §469,  formula  [10].) 


ha=    Vs{s  —  a){s  —  h){s  —  c) 

0/ 


he  =   Vs{s  —  a){s  —  b)  {s — c) 


(1) 

(2) 
(3) 


How  can  (2)  and  (3)  be  obtained  from  (1)  by  analogy  ? 

•The  law  of  formula  (11)  was  introduced  into  mathematical 
texts  by  Heron  of  Alexandria  in  the  first  century  B.C. 


AREAS.    LITERAL  EQUATIONS.    FACTORING     313 

EXERCISES 

1.  In  the  triangle  ABG,  a=10,  6  =  17,  c  =  21.     Find  ha. 


2 


ha  =  -Vs{s—a){s  —  b){s—c) 

5  =  |(a+6+c)  =1(10+17+21)  =24 
s  — a  =  14,  s— 6  =  7,  s— c  =  3. 

Substitute  these  values  in  the  formula,  and 


2    ,„.__,     1 


^a  =  jQl/24  •  14  •  7  •  3  =  5^  4  •3-2-2-7-7-3 

1     1  84         4 

=  ^v/4-9-4-49  =  ^(2  •  3  •  2  •  7)  =-^-  =  16^  . 

2.  Find  the  altitudes  of  each  of  the  following  triangles: 

(1)  a  =  35,  6  =  29,  c=  8 

(2)  a  =  70,  6  =  65,  c=  9 
1(3)  a  =  45,  6  =  40,  c=13 

3.  The  sides  of  a  quadrilateral  are  as  follows : 

AB  =  29,  BC  =  S,  CD  =  28,  DA  =  21,  and  the  diagonal  AC  =  30. 
Find  the  area  and  the  distance  from  D  to  AC. 

471.  Area  of  an  equilateral  triangle.     The  area  of  an 
equilateral  triangle  is  one-fourth  the  square  of  a  side  times 

the  square  root  of  3,  or,  in  symbols,  A  =  ^1/3  . 

The  area  of  triangle  ABC,  Fig.  388,  is  given  by  the  formula 
AABC  =  \ah 

Show  that  /i2  =  ^2  -  y  =  — 

4      4 

By  substitution,    l\ABC  =  \a  •  -Vs 
:.     AABC=^VZ 


314  SECOND-YEAR  MATHEMATICS 

EXERCISES 

1.  Find  the  areas  of  the  following  equilateral  triangles,  hav- 
ing the  side  equal  to  12;   10;  4;  8;  c-\-d;  2mn. 

2.  Find  the  side  of  an  equilateral  triangle  whose  area  is 

251^3;  101^3. 

In  proving  the  formula  for  the  area  of  a  triangle  in 
terms  of  the  sides,  §  469,  we  have  factored  the  polynomials 
2ab-\-¥-c''-\-a^  and  2ah-¥+c^-a'^. 

In  §  472-476  we  shall  study  further  the  method  used 
in  factoring  these  polynomials  as  well  as  some  other  fre- 
quently occurring  polynomial  forms. 


Polynomials  Factored  by  Grouping 

472.  The  terms  of  some  polynomials  may  be  grouped 
to  show  a  common  binomial  factor. 

1.  Factor  3a+36+5fl+5& 

Grouping  the  first  two  terms  and  the  last  two  terms, 

3a+36+5a+56  =  3(a4-&)+5(a+6)  =  (a+6)(3+5) 

Test  by  multiplication. 

2.  Factor  ac+hc-{-ad-\-bd 

ac+bc+ad+bd  =  c(a+b)+d{a-\-b)  =  ia+b){c+d) 

Test  by  multiplication. 

3.  Factor  lix^-Qx'^-21x+9 

14a:'-6xa-21x+9  =  2x2(7x-3)-3(7x-3)  =  (7a;-3)(2x«-3) 
Test  by  substitution  and  by  multiplication. 


AREAS.    LITERAL  EQUATIONS.    FACTORING     315 


EXERCISES 


Resolve  into  factors  the  following  expressions  and  test 
results,  doing  as  many  as  you  can  mentally: 


1.  ax-{-bx-\-am-{-bm 

2.  ar-\-br-\-as-\-bs 

3.  ad+bd+av+bt 
4t.  Sa+3b+ay+by 
6.  ak—bk-\-al—bl 

6.  ax^—bx^+ay^—by^ 

7.  abc-{-abx-\-nc-\-nx 

8.  a^k-\-aH-{-¥k+bH 

9.  5au^5av-\-mu—mv 


16.  9-15r+27r2-45r3 

17.  8gh+12ah+l0bg+15ab 

18.  15z-Q-20zw+8w 

19.  2m^+Skm-Umn-2lkn 

20.  Sax+3ab-^2x^+2bx-\-b-\-x 

21.  4x^-\-^-4xh-^ 

22.  l-\-r—r'^xy—r^xy 

23.  x'^-x^+1-x 

24.  (a+m)(c+n)-2n(a+m) 


10.  w^a+ma^+m^a^+m^a^    26.  {x+y)(ai-b)-{x+y){b+c) 


11.  a2_o^-f-a6-6(i 

12.  x6+5a;4+x34-5x 

13.  Qx'^-9x-l0xy+15y 

14.  2m3+m2H-6m+3 

15.  Sac+3ax  —  5c  —  5x 


26.  m(x+?/)2+(a;+?/) 

27.  a2(2a+l)2-2a-l 

28.  a—b+a^y—b^y 

29.  (c+d)(c2+d2)+2c2d+2cd2 

30.  (x+yy{x-y)-{x-yy{x-{-y) 


Reduce  the  following  fractions  to  lowest  terms: 


31. 


32. 


ax-{-bx+am-\-bm, 
ar-]-br-{-as-{-bs 

SuSv-[-au—av 
5bu —5bv+2ku— 2kv 


33^    ax^-bx^-i-ay^-by^ 
mx^-i-my^-\-nx^-}-ny'^ 


34. 


x*-2x^+7x-U 
"2x3-4x2+6x-12 


316  SECOND-YEAR  MATHEMATICS 

473.  The  terms  of  some  polynomials  may  be  grouped 
to  show  the  difference  of  two  squares. 

EXERCISES 

Factor  the  following  polynomials: 

1.  a^-2ab+¥-c'' 

Grouping  the  first  three  terms,  a^—2ab-^h^  —  c'^  equals 
(a-6)2-c2  =  (a-6+c)(a-6-c) 

2.  x^-Qxy+9y^-lQz'^  5.  l-a^-2ab-¥ 

3.  25x2+16?/2-4a2+40x?/  6.  9m'^-a'^-'iab-4:¥ 

4.  ¥-x''-2xy-y''  7.  36r2-4+20^-25^2 

8.  x'^+2xy^y'^-a''-2ab-¥ 

9.  a2+2a+26c-62-c2+l 

10.  9x2+16?/2-49a2-462-f  28a6+24a:?/ 

11.  9a2-12a6+462-16x2-8a:?/-?/2 

474.  The  terms  of  some  polynomials  can  be  grouped 
to  show  a  perfect  square. 

EXERCISES 

Factor  the  following  polynomials: 

1.  a''+2ab+¥-\-Qa+Qb+9 

Grouping  the  first  three  terms,  the  4th  and  5th  terms,  and  keep- 
ing the  last  term  separate,  we  have, 
o2+2a6+62+6a+66+9  =  (a+6)2+6(a+6)+9  =  (a+6+3)2 

2.  m2+2mnH-w2+2m+2n+l 

3.  m2+2mn+n2+6am+6an+9a2 

4.  a^+¥+c^-[-2ab-^2ac-^2bc 


AREAS.     LITERAL  EQUATIONS.     FACTORING     317 

475.  The  terms  of  some  polynomials  may  he  grouped 
to  show  a  trinomial  which  can  he  factored  hy  the  trial  method. 

EXERCISES 

Factor  the  following: 

1.  x^-{-y'^-\-z—2xy  —  y—Q 

Grouping  the  first,  second,  and  fourth  terms,  the  third  and 
fifth  terms,  and  keeping  the  sixth  term  separate, 

x^i-if+x-2xy-y-Q  =  x^-2xy+y^-{-x-y-Q 
=  {x-yy  +  {x-y)-Q  =  {x-y+3)ix-y-\-2) 

2.  a2+2a6+62+3a+36-10 

3.  a2-6o6H-962+7ac-216c-44c2 

4.  m*x+m^x  —  650x 

6.  4x^-\-Sxy+^y^-{-13xi-13yi-S 
6.  3c2-6cd+3(i2-2c+2(i-5 

476.  Some  trinomials  may  he  factored  hy  first  changing 
them  to  complete  squares. 

EXERCISES 

Factor  the  following  trinomials : 

1.  x^-\-xY+y^ 

By  adding  x'^y'^  to  the  trinomial  x^-{-x'^y^+y*,  it  becomes  a  per- 
fect square:  x*-\-2xY-\-y*.  However,  this  changes  the  value  of 
the  trinomial.  To  keep  the  value  unchanged  x^  is  subtracted  from 
the  trinomial.  Thus,  x*+xY+7/  =  x*+2x'^y^+y*—xY.  This  may 
be  written:  (x^-{-y^y  —  (xy)^.  This  is  the  difference  of  two  squares 
and  its  factors  are  ix^-{-y^+xy){x^-{-y^—xy). 

2.  a'^-7a''¥+¥  5.  25x^+31xY+l^y* 

3.  x'+x'^+l  6.  a^x^+a^x^+a'' 

4.  lQx^-17xY+y*  7.  4:9a'¥-53a%''x^+^x* 


318  SECOND-YEAR  MATHEMATICS 

8.  9x*-10xY+y^  9-  4a4-5a262+64 

The  difference  of  two  squares  may  be  obtained  in  exercise  9  by 
adding  and  subtracting  either  a%^,  giving  4ia*~4:a^b^-\-b*—a'^h'^,  or 
by  adding  and  subtracting  9a%^,  giving  4a^+4a262-f-64_  9^252 

Show  that  both  lead  to  the  same  prime  factors.  Show  also  that 
exercise  8  gives  two  pairsof  factors  that  lead  to  the  same  prime  factors. 

10.  Factor  the  following  trinomials  by  adding  and  subtract- 
ing a  monomial  square: 

1.  a;^+4 

Add  and  subtract  4x^ 

2.  4x^+1  5.  a4+32464 

3.  m4+4  6.  1024x*+y^ 

4.  a'b^-\-Q4:  7.  81x*+^y^ 

477.  Summary  of  factoring.  Polynomials  to  be 
factored  may  be  classified  according  to  the  number  of 
terms  they  contain. 

I.  If  the  polynomial  is  a  binomial  it  may  be  of  the 
following  types : 

1.  The  difference  of  two  squares,  as  x^—y^.  The 
factors  are  {x-\ry){x  —  y). 

2.  The  difference  of  two  cubes,  as  x^  — ^/^     The  factors 

are  {x  —  y){x^-\-xy-[-y'^). 

3.  The  sum  of  two  cubes,  as  7?-\-y^.    The  factors  are 

{x-\-y){x'^-xy-^y'^). 

11.  If  a  polynomial  is  a  trinomial  it  may  be  of  the 
following  types: 

1.  The  perfect  square,  as  x^  =*=  2xy-\-y'^.    The  factors  are 


AREAS.    LITERAL  EQUATIONS.    FACTORING     319 

2.  A  trinomial  which  may  be  changed  into  a  perfect 
square  by  adding  a  term,  as  x^+x^y^+y"^.  This  is  changed 
to  a;'*+2a:VH-2/^  — a^y  and  is  then  factored  as  the  difference 
of  two  squares. 

3.  A  trinomial  of  the  form  ax'^-^hx-\-c.  Such  tri- 
nomials may  be  factorable,  having  factors  obtainable  by 
the  trial  method. 

IIL  Polynomials  not  of  any  of  the  types  in  I  and  II 
may  be  factored : 

1.  By  dividing  each  term  by  a  common  factor,  as 
ax+ay.     The  factors  are  a{x-{-y). 

2.  By  grouping  its  terms  so  as  to  change  it  to  the  form 
of  one  of  the  preceding  types.  Thus,  ax-\-hx-\-ay-{-hy 
when  grouped,  takes  the  form  {ax-]-hx)-\-{ay-\-hy).  This 
equals  x{a-{-b) -\-y{a-{-h) ,  which  is  of  type  III,  1. 

Similarly,  the  polynomial  x^-\-2xy-^y^ — a^  —  2ab  —  b^  is 
changed  to  x^+2xy-{-y'^—{a^-{-2ab-\-b-),  which  is  of  type 
1,2. 

Miscellaneous  Review  of  Factoring 

478.  Factor  the  following  polynomials: 

1.  2Qxyz-\-Q5xy'^  9.  z~—x'^-\-2xy—y'^ 

2.  7x4-35x2+140:3  10.  a''-8ab-\-15b'' 

3.  32 -16a+ 1862 -962a  11.  7n''-4mn-77n'^ 

4.  m2+|m/i-4mp-3n/7         12.  343a3+12563 

6.  121m2n2-64p2g2  13.  a6+4a-36-12 

6.  SlxY-2*  14.  a:2+22x+121 

7.  32mn4-162m  15.  x'^+x'^-x-1 

8.  lQx^+^9y^-5Qxy  16.  (m-n)2-ll(m~n)-12 


320  SECOND-YEAR  MATHEMATICS 

17.  a;3-343 

18.  5a+6a2+l 

19.  56-15a+a2 

20.  xY+^0xy+104c 

21.  Sxh/z^-lSz* 

22.  8a;9+729 


23.  a6_|_25a3+24 

24.  ?7z8-38m4+105 


25.  64-a;6 

26.  9a;2-[-24x2/+16i/2 

27.  362-146a+8a2 

28.  3m2+4(2m+l) 

29.  x''-\-if-{-2xy-a'-b''-2ah 

30.  m'^+f+2mt-x^-if-2xy 

31.  25a4-26a262-f64  (2  pairs) 

32.  4x4-13xy+V  (2  pairs) 


Summary 

479.  The  following  theorems  were  proved  in  this 
chapter: 

1.  Parallelograms  having  equal  bases  and  equal  altitudes 
are  equal. 

2.  A  parallelogram  is  equal  to  a  rectangle  having  the 
same  base  and  altitude. 

3.  A  triangle  is  equal  to  one-half  a  parallelogram  having 
the  same  base  and  altitude. 

4.  The  square  on  the  hypotenuse  of  a  right  triangle  is 
equal  to  the  sum  of  the  squares  on  the  sides  including  the 
right  angle. 

5.  In  a  triangle  the  sum  of  the  squares  of  two  sides  is 
equal  to  twice  the  square  of  one-half  of  the  third  side  increased 
by  twice  the  square  of  the  median  to  the  third  side. 

6.  The  area  of  a  triangle  is  equal  to  one-half  the  product 
of  the  base  and  altitude, 


=A: 


AREAS.    LITERAL  EQUATIONS.    FACTORING     321 

7.  The  area  of  a  triangle  is  equal  to  one-half  the  product 
of  two  sides  by  the  sine  of  the  included  angle, 

A  =  \ab  sin  C. 

8.  The  area  of  a  triangle  is  equal  to  one-half  the  perim^ 
eter  times  the  radius  of  the  inscribed  circle,  

A=\p.r. 

9.  The  area  of  a  triangle  is  equal  to  the  product  of  the 
three  sides  divided  by  fowp-  times  the  radius  of  the  circum- 
scribed  circle, 

.    abc 

10.  The  area  of  a  triangle  is  equal  to 


A  =  vs{s-a){s-b){s-c). 


11.  The  area  of  an  equilateral  triangle  is  one-fourth  the 
square  of  a  side  times  the  square  root  of  3 

480.  The  chapter  has  given  drill  in  solving  literal 
equations  in  one  and  two  unknowns  and  in  factoring  poly- 
nomials. 


CHAPTER  XIX 

AREAS  OF  POLYGONS.    AREA  OF  THE  CIRCLE. 
PROPORTIONALITY  OF  AREAS 

Areas  of  Polygons 

481.  Area  of  the  rectangle.  The  rectangle  is  the 
fundamental  figure  by  which  the  areas  of  all  other  recti- 
linear figures  are  measured.  In  the  first-year  course  we 
have  seen  that  the  area  of  the  rectangle  is  given  by  the 
formula 

S  denoting  the  area,  h  the  base,  and  h  the  altitude.     In  the 
form  of  a  theorem  this  is  stated  as  follows : 

The  area  of  a  rectangle  is  equal  to  the  product  of  the  base 
by  the  altitude. 

»  The  formula,  S  =  b  '  h,  which  was  shown  to  hold  foi 
rational  values  of  b  and  h,  is  also  true  when  b  and  h  are 
irrational.    This  may  be  shown  as  follows: 

Let  6=^12  =  3.464101 and  /i=v^ 27  =  5. 196152 

Then  the  following  table  gives  the  areas  of  rectangles, 
the  lengths  of  whose  sides  vary,  being  approximations 


Rectangle 

b 

h 

S  =  b.h 

I 

3.464 

5.196 

17.998944 

II 

3.4641 

5.1961 

17.99981001 

Ill 

3.46410 

5.19615 

17.999983215 

IV 

3.464101 

5.196152 

17.999995339352 

of  l/l2  and  1^27  to  three,  four,  five,  and  six  decimal 
places  and,  therefore,  rational  numbers.  Hence,  the 
formula  S  =  b  '  h  may  be  applied  in  each  case. 

322 


03 

.a 

02 

i 

■^ 
ft 

73 


FOLYGONS.    CIRCLES.    PROPORTIONALITY      323 


It  is  seen  from  the  table  that  the  difference  between  18 
and  the  several  areas,  I,  II,  III,  and  IV  decreases,  being 
less  than  .002,  .0002,  .00002,  .000005,  respectively.  By 
taking  h  and  /i  to  a  greater  number  of  decimal  places,  this 
difference  will  continue  to  decrease,  in  fact  it  can  be  made 
less  than  any  assigned  number,  however  small.  The  area 
is  accordingly  said  to  approach  18  as  a  limit.  The  same 
result  is  obtained  by  applying  the  formula 

S  =  h  •  /i  =  /l2  •  l/27=v'22  .  3  .  33  =  18. 

482.  Theorem:  The  area  of  a  parallelogram  is  equal 
to  the  product  of  the  base  and  altitude.     Prove.  Use  §  460. 

483.  Theorem:  The  area  of  a  trapezoid  is  equal  to 
one-half  the  product  of  the  altitude  by  the  sum  of  the  bases. 
Prove  (see  Fig.  389). 

Show  that  the  area  of  a 
trapezoid  is  equal  to  the  prod- 
uct of  the  altitude  by  the 
median  (see  §  161). 

484.  Theorem:  The  area  of  a 
regular  inscribed  polygon  is  equal  to 
the  product  of  one-half  of  the  perim- 
eter and  the  perpendicular  from  the 
center  to  the  side  (apothem). 

Draw  AO,  BO ,  Fig.  390. 

Denote  the  length  of  a  side  of 
the  polygon  by  a,  the  perpendicular 
from  the  center  to  the  side  by  h, 
the  number  of  sides  by  n,   i^ 

Then,  AAOB=^, 


Fig.  389 


Fig.  390 


2 
ABOC  =  ^,etc. 


Why 


ABCD.. 


nah  _p  •  h 


Why? 


324  SECOND-YEAR  MATHEMATICS 

485.  Theorem:  The  area  of  a  regular  circumscribed 
polygon  is  equal  to  the  product  of  one-half  the  perimeter  and 
the  radius. 

Show,  Fig.  391,  that  .^-^'^^'-v^ 

AAOB  =  ^,  f^  \ 

ABOC  =  ^,  etc.  ^4         / i^^x" / ^ 


ABCD...='^  =  ^ 


2         2  .-  ^ 

Fig.  391 

EXERCISES 

1.  Express  in  terms  of  the  radius  the  areas  of  the  inscribed 
and  circumscribed  squares  (see  exercises  1.,  2,  §  440). 

2.  The  area  of  a  square  is  16  square  centimeters.  Find 
the  diameters  of  the  inscribed  and  circumscribed  circles. 

3.  Prove  that  the  area  of  the  equilateral  inscribed  triangle 

is  |r-V3  (see  exercise  9,  §  441). 

Thus,  the  area  of  the  equilateral  inscribed  triangle  varies  as 
the  square  of  the  radius.     Give  reason. 

4.  Prove  that  the  area  of  the  circumscribed  equilateral 
triangle  is  Sr^i^S  . 

Show  that  the  area  varies  as  the  square  of  the  radius.  Show 
that  the  area  is  a  function  of  the  radius.  Sketch  freehand,  without 
plotting  points,  the  graph  of  this  function. 

5.  Prove  that  the  area  of  the  regular  inscribed  hexagon  is 

6.  Prove  that  the  area  of  the  circumscribed  regular  hexagon 
is2r2v/3. 

7.  Find  the  area  of  a  regular  hexagon  whose  side  is  6  inches. 

8.  The  radius  of  a  circle  is  10.  Find  the  area  of  the  inscribed 
regular  hexagon. 


POLYGONS.    CIRCLES.    PROPORTIONALITY      325 


9.  The  diameter  of  a  circle  is  8.  Find  the  area  of  the  regular 
inscribed  hexagon. 

10.  Prove  that  in  the  same  circle  the  area  of  the  regular 
inscribed  hexagon  is  twice  as  large  as  that  of  the  equilateral 
inscribed  triangle. 

486.  Area  of  any  polygon.  The  areas  of  polygons 
may  be  found  by  dividing  the  polygons  into  triangles,  as 


Fig.  392  Fig.  393 

in  Fig.  392,  or  into  triangles  and  trapezoids,  as  in  Fig.  393. 

Area  of  the  Circle 

487.  If  the  midpoints  of  the  arcs  subtended  by  the 
sides  of  a  given  regular  inscribed  polygon,  as  triangle 
ABCy  Fig.  394,  are  joined  to  the 
adjacent  vertices  of  the  polygon, 
a  regular  inscribed  polygon, 
AFBECD,  is  formed  having  twice 
as  many  sides  as  the  given  polygon 
(see  §436). 

The  perimeter  of  the  second 
polygon  is  greater  than  that  of  the 
first.     Why? 

If  the  process  of  doubling  the 
number  of  sides  is  continued,  the  perimeter  increases  as 
the  number  of  sides  increases.  It  can  be  made  to  differ 
from  the  length  of  the  circle  by  less  than  any  quantity, 
however  small.  The  perimeter  is  said  to  approach  the 
circle  as  a  limit. 


326  SECOND-YEAR  MATHEMATICS 

The  apothem  OX  approaches  the  radius  as  a  limit. 
The  area  of  the  polygon  approaches  the  area  of  the 
circle  as  a  limit. 

488.  If  tangents  are  drawn  at  the  midpoints  of  the 
arcs  terminated  by  consecutive  points  of  contact  of  the 
sides  of  a  given  regular  circum- 
scribed polygon,  as  ABCD,  Fig.  395, 
a  regular  circumscribed  polygon,  as 
EFGHIKLM  is  formed  having 
twice  as  many  sides  as  the  given 
polygon  (see  §438). 

The  perimeter  of  the  second 
polygon  is  less  than  that  of  the 
first.     Why? 

If  the  process  of  doubling  the 
number  of  sides  is  continued,  the  perimeter  decreases  as 
the  number  of  sides  increases.  It  can  be  made  to  differ 
from  the  length  of  the  circle  by  less  than  any  quantity, 
however  small,  thus  approaching  the  circle  as  a  limit. 

The  area  of  the  polygon  approaches  the  area  of  the 
circle  as  a  limit. 

489.  According  to  §§487  and  488,  the  area  of  the 
circle  is  the  common  limit  approached  by  the  areas  of 
the  inscribed  and  circumscribed  regular  polygons,  as  the 
number  of  sides  increases  indefinitely. 

These  areas  are  given  by  the  formulas : 

7)h         Pr 

^  and  -^ ,  respectively  (see  §§  484,  485). 

As  the  number  of  sides  of  the  polygons  is  increased 

7)h  c  *  r 

indefinitely,  —-  approaches  — ^  as   a   limit,    for   p  ap- 
proaches c,  and  h  approaches  r. 

Pr  c  *  r 

-^  approaches  -^  as  a  limit,  for  P  approaches  c. 


POLYGONS.    CIRCLES.    PROPORTIONALITY      327 


CT 

Hence,  the  common  limiting  value,  ^ ,  expresses  the 

area  of  the  circle. 

In  words,  this  may  be  stated  as  follows : 

Theorem:  The  area  of  a  circle  is  one-half  the  product 
of  the  length  of  the  circle  and  the  radius,  i.e.,  area  of  circle 
is  given  by 

Since,  c  =  27rr,  it  follows  that  the  area  of  a  circle  is 
given  by 

Trr\ 


Show  that  the  area  of  a  circle  is  a  function  of  the  radius 
and  sketch  the  graph  of  this  function. 

490.  Theorem:  The  area  of  a  sector  of  a  circle  is  equal 
to  one-half  the  product  of  the  radius  and  the  length  of  the 
arc  of  the  sector. 

We  have  seen  in  §  297  that  central  angles  have  the 
same  measure  as  the  intercepted  arcs  and  that  two  central 
angles  are  to  each  other  as  the  intercepted  arcs  (§  297, 
exercise  8). 

Hence,  |  =  p,Fig.  396. 

Similarly,  we  may  show  that  equal 

central  angles  include  equal  sectors  and 

that  two  sectors  are  to  each  other  as  their 

central  angles. 

a    A 
Hence,  h^^  Fig.  396 

*  A  proof  of  the  theorem  is  not  attempted,  as  this  is  considered 
beyond  the  province  of  secondary-school  work. 


328  SECOND-YEAR  MATHEMATICS 

Denote  by  a  the  number  of  degrees  in  a  central 
angle,  and  consider  the  circle  as  an  arc  whose  central 
angle  is  360°. 

rpu         a        a'  ,    ,    irra 

^^"^'360  =  2^r'^^^^=i80 (^) 

Similarly,  A=_^  Why? 

360  "2     180 
or  A  =  la' '  r (5) 


^  =  Qm=i-TQn-^  Why? 


491.  Area  of  a  segment.  The  area  of  a  segment 
ACB,  Fig.  397,  may  be  found  by  sub- 
tracting the  area  of  triangle,  AOB,  from 
the  area  of  the  sector,  AOBC,  the  area 
of  triangle  AOB  being  computed  by 
means    of    the    formula    T  =  ^^   sin  0, 

§  466;  or,  by  T  =  ia  yjr'-^ ,  §  233. 

Hence  the  area  of  a  segment  is  given  ^^^-  ^^^ 

by  the  following  formulas : 

(1)  S  =  ^a'r  —  ^r'^  sin  X,  where  X  is  the  central 
angle  subtended  by  the  chord  a. 

Orhy  (2)  S  =  ^a'r-ia.yJ^^. 

Where  a  is  the  length  of  the  chord,  a'  the  length  of 
the  arc,  and  r  the  radius  of  the  circle. 

EXERCISES 

1.  The  area  of  a  circle  is   64.     Find   the  diameter  and 
length. 

2.  Find  the  diameter  of  a  circle  whose  area  is  1  square  inch; 
1  square  foot;  1  square  yard. 


POLYGONS.    CIRCLES.    PROPORTIONALITY      329 

3.  What  is  the  area  of  the  ring  formed  by  two  concentric 
circles,  Fig.  398,  whose  radii  are  5  inches  and 
6  inches,  respectively;    a  inches  and  h  inches, 
respectively  ? 

4.  The  length  of  a  circle  is  50  inches.    What 
is  the  area  ? 

Fig.  398 

5.  The  area  of  a  circle  is  616  square  inches. 

How  many  degrees  are  there  in  an  angle  at  the  center  that 
intercepts  an  arc  11  inches  long? 

6.  The  radius  of  a  circle  is  100  feet.  The  length  of  the  arc 
of  a  sector  is  25  feet.    Find  the  area  of  the  sector. 

Use  formula,  (J5),  §  490. 

7.  The  radius  of  a  sector  is  9  inches,  its  area  is  72  square 
inches.    Find  the  length  of  the  arc. 

8.  The  area  of  a  sector  is  a  square  foot,  and  the  radius  is  r  feet 
long.    Find  the  length  of  the  arc. 

9.  The  radius  of  a  circle  is  8  inches.  Find  the  area  of  a  sector 
with  arc  36°. 

Make  use  of  the  fact  that  the  area  of  the  sector  is  tV  of  the  area 
of  the  circle. 

10.  Find  the  area  of  the  segment  whose  arc  is  36°  in  a  circle 
of  radius  12  inches. 

When  finding  the  area  of  the  triangle  notice  that  the  base  of  the 
triangle  is  the  side  of  a  regular  10-side,  exercise  5,  §  443,  or  use 
the  formula  \ah  sin  C. 

11.  Find  the  area  of  a  segment  of  arc  72°,  in  a  circle  of 
radius  20. 

12.  The  area  of  a  circle  is  15,400  square  inches.  Find  the  area 
of  a  segment  whose  arc  is  60°. 


330  SECOND-YEAR  MATHEMATICS 

Proportionality  of  Areas 

The  proofs  of  the  theorems  in  §§  492-496  are  very 
simple  and  are  left  to  the  student. 

492.  Theorem:  Two  parallelograms  are  to  each  other 
as  the  products  of  their  bases  and  altitudes,  i.e.,  -p/^jrrn  • 

493.  Theorem:  Two  parallelograms  having  equal  bases 
are  to  each  other  as  the  altitudes,  i.e.,  -^  =  1-  - 

By  alternation,  t^ = t^  • 

til       rl2 

Thus,  if  the  base  of  a  parallelogram  remains  fixed  and  if  the 
altitude  varies  continuously,  taking  successive  values  hi,  h^, 

hz, ,  etc.,  P  takes  the  corresponding  values  Pi,  P2, 

p 
P3, ,  etc.    However,  -7-  remams  constant,  i.e., 

P}=P}=P.'= ,etc 

hi     hi     hz 

Denoting  this  constant  ratio  by  b,  we  have  Pi  =  bhi,  P2=bh2, 
Pz  =  bhz,  etc.  Show  that  P  is  a  function  of  h.  Without  plotting 
points,  sketch  the  graph  of  this  function. 

Hence,  P  is  directly  proportional  to  h,  or  P  varies  directly 
as  h  if  the  base  b  remains  constant. 

494.  Theorem:  Two  triangles  are  to  each  other  as  the 
products  of  the  bases  and  altitudes. 

495.  Theorem:  Areas  of  triangles  having  equal  bases 
are  to  each  other  as  the  altitudes. 

496.  Theorem:  Areas  of  triangles  having  equal  alti- 
tudes are  to  each  other  as  the  bases. 


POLYGONS.    CIRCLES.    PROPORTIONALITY      331 


EXERCISES 

1.  Show  that  the  area  of  a  triangle  having  a  fixed  base  varies 

T 
directly  as  the  altitude,  i.e.,  show  that  -j-  remains  constant,  as  h 

varies. 

2.  Show  that  the  area  of  an  equilateral  triangle  varies 
directly  as  the  square  of  the  side. 

3.  Show  that  the  area  of  a  circle  varies  directly  as  the  square 
of  the  radius. 

497.  Theorem:  The  areas  of  two  triangles  that  have 
an  angle  in  one  equal  to  an  angle  in  the  other  are  to  each 
other  as  the  products  of  the  sides  including  the  equal  angles. 

Given   AABC  and  A'B'C  having  C=  C,  Fig.  399. 
To  prove  that    Y'^afh'' 


Proof:    T  =  \ah  sin  C. 


T_ 

r 


ah 


Why? 

Why? 

Why? 


EXERCISES 

1.  Two  triangles  have  an  angle  in  each  equal.  The  includ- 
ing sides  of  one  are  48  and  75,  those  of  the  other  triangle  are 
and  45  and  70.    Find  the  relative  areas  of  the  triangles. 

2.  Two  sides  of  a  triangular  building  are  150  ft.  and  130 
feet.  Wliat  part  of  the  whole  building  is  included  by  50  ft. 
on  the  first  side  and  30  ft.  on  the  second  ? 


332  SECOND-YEAR  MATHEMATICS 

3.  A  triangular  lot  extends  60  ft.  and  80  ft.  on  two  sides 
from  a  corner.  If  a  building  is  to  front  50  ft.  on  the  first  side, 
how  many  feet  on  the  second  side  should  it  occupy  to  cover 
J  of  the  lot? 

4.  Two  sides,  a  and  b,  of  a  triangle  are  9  and  15  respectively. 
ShBw  where  a  line  going  through  a  point  on  a  and  5  units  from 
the  common  vertex  of  a  and  b  must  intersect  the  side  b  to  bisect 
the  surface  of  the  triangle. 

498.  Theorem:  The  areas  of  similar  triangles  are  to 
each  other  as  the  squares  of  the  homologous  sides. 


Show  that  Y'^Vh'^Vh"  ^^^'  ^^^' 


h^b 
h'    b'' 

r  b''b' 


EXERCISES 


Why? 


1.  The  side  of  a  triangle  is  10  inches.     Find  the  correspond- 
ing side  of  a  similar  triangle  having  twice  the  area. 

2.  Two  similar  triangles  have  two  homologous  sides  5  and 
15  respectively.    What  is  the  ratio  of  the  areas  ? 

3.  Bisect  the  surface  of  a  triangle  by  a  line  dr9,wu  from  a 
vertex  to  the  opposite  side, 


POLYGONS.    CIRCLES.    PROPORTIONALITY      333 

499.  Theorem:    The  areas  of  similar  polygons  are  to 
each  other  as  the  squares  of  the  homologous  sides. 

Given  polygon  ABC ^  polygon  A'B'C 

(Fig.  401).     Let  P  denote  the  area  of  ABC and 

P'  denote  the  area  of  A'B'C 


To  prove 


P_^d^ 
P'      d'^ ' 


Fig.  401 


Proof:  Divide  ABC and  A'B'C into 

triangles  J,  7/,  III,  etc.,  and  F,  IP,  III',  etc.,  respectively, 
by  drawing  diagonals  from  homologous  vertices  as  B 
and  B', 

Then  I^  P,  11^  IP,  etc. 

•*•     P~c'^'  IP~d'^" 


Show  that 


d2 


,  etc. 
, . ,  etc. 


Why? 
Why? 


"    P    IP    IIP 


etc. 


Why? 


Why? 


I  +  II  +  III+ ^n^d^ 

P+IP-\-IIP-{- IP    d'^  ^^     ^' 


P'    d'2 


334 


SECOND-YEAR  MATHEMATICS 


EXERCISES 

1.  Two  homologous  sides  of  two  similar  triangles  are  5  and 
8.    The  area  of  the  first  is  150.     Find  the  area  of  the  second. 

2.  If  one  square  is  9  times  as  large  as  another,  what  is  the 
relative  length  of  the  homologous  sides  ? 

3.  The  area  of  a  polygon  is  6j  times  the  area  of  a  similar 
polygon.  A  side  of  the  smaller  is  4  feet.  Find  the  length  of  the 
homologous  side  of  the  larger. 

4.  Show  that  if  equilateral  triangles  are  constructed  on  the 
sides  of  a  right  triangle,  the  triangle  on  the  hypotenuse  is  equal 
to  the  sum  of  the  triangles  on  the  other  two  sides. 

5.  Show  that  if  semicircles  are  drawn  on  the  sides  of  a  right 
triangle,  the  area  of  the  semicircle  on  the  hypotenuse  is  equal 
to  the  sum  of  the  areas  of  the  semicircles 
on  the  two  sides  of  the  right  angle. 

J6.  Semicircles  are  drawn  on  the  sides 
of  a  right  triangle,  Fig.  402.  Show  that 
the  sum  of  the  areas  of  lunes  I  and  II  is 
equal  to  the  area  of  the  right  triangle 
(theorem  of  Hippocrates,  430  b.c). 

7.  Similar  polygons,  Pi,  P2,  and  P3, 
are  drawn  on  the  sides  of  a  right  triangle 
as  homologous  sides.  Fig.  403.  Prove 
that  P3,  the  area  of  the  polygon  on  the 
hypotenuse,  is  equal  to  the  sum  of  Pi 
and  P2. 

P2^62 
Pz      &' 


Fig.  403 


Proof: 


Why? 
Why? 


P1+P2    a2+62 


Why 


P3      ^     c2     • 

"  "  J    • 

(Pi+P2)c2  =  P3(a2+62). 

Why? 

Pl+P2  =  P3. 

Why? 

POLYGONS.    CIRCLES.    PROPORTIONALITY      335 

8.  The  homologous  sides  of  similar  hexagons  are  9  in.  and 
12  in.,  respectively.  Find  the  homologous  side  of  a  similar 
hexagon  equal  to  their  sum. 

500.  Theorem:  The  areas  of  two  circles  are  to  each 
other  as  the  squares  of  the  radii,  or  as  the  squares  of  the 
diameters. 

EXERCISES 

1.  What  is  the  ratio  of  the  areas  of  two  circles  whose  radii 
are  5  in.  and  10  inches  ? 

2.  The  areas  of  two  circles  are  in  the  ratio  2  to  4.  What  is 
the  ratio  of  the  diameters  ? 

3.  The  radii  of  two  circles  are  to  each  other  as  3 : 5,  and  their 
combined  area  is  3850.    Find  the  radii  of  the  two  circlea. 

4.  The  radii  of  two  circles  are  to  each  other  as  7 :  24,  and  the 
radius  of  a  circle  whose  area  is  equal  to  their  sum  is  50.  Find 
the  radii  of  the  first  two  circles. 

Problems  of  Construction 

501.  Make  the  following  constructions. 


1.  Construct  a  square 
equal  to  the  sum  of  two  or 
more  given  squares. 

Given  x,  y,  z,  w,  the  sides 
of  given  squares. 

Required  to  construct  a 
square  equal  to  the  sum  of  the 
given  squares.  Fig.  404  suggests 
the  construction. 

Prove  that 


Fig.  404 


2.  Construct  a  square  equal  to  four  times  a  given  square. 


336 


SECOND-YEAR  MATHEMATICS 


D    i     ^   1 


I    A 
Fig.  405 


3.  Construct   the   square  root  of  an 
integral  number. 

Make    the    construction,    Fig.'  405,   on 
squared  paper. 

Measure  AC,  AD,  AE,  AF  and  check  by 
extracting  the  square  roots   of   2,  3,  4,  5. 

4.  Transform  a  polygon  into  a  triangle 
equal  to  it. 

Draw  the  diagonal  AD, 
Fig.  406. 

Through  E  draw  EF  \\  AD 
intersecting  the  extension  of 
AB  in  F. 

Draw  DF  and  show  that 
ADFA  =  ADEA. 

Show  that  FBCD  is  equal 
to  ABODE. 

This  reduces  the  pentag-^n  to  the  equivalent  quadrilateral  FBCD. 

Draw  the  diagonal  DB. 

Draw  CG  ||  DB. 

Draw  DG. 

Show  ADCB  =  ADGB. 

Show  that  FBCD  =  AFGD,  which  is  the  required  triangle. 

.-.    ABCDE  =  AFGD.        Why? 

5.  Draw  a  square  equal  to  a  given  triangle. 

Analysis :  Since  the  area  of  the  triangle  is  ^bh  and  since  the  area 
of  the  square  is  a^,  we  must  have  a^  =  ^bh,  where  b  and  h  are  known, 
and  a  unknown.  Hence,  the  problem  reduces  to  constructing  the 
mean  proportional  between  ^b  and  h. 

E 


Fig.  406 


Construction:  On  AB,  Fig.  407,  lay  off  AC  =  \b  and  CD  =  /i. 
Draw  QEl^AD, 


POLYGONS.    CIRCLES.    PROPORTIONALITY      337 

Draw  the  semicircle  on  AD. 

Draw  a  square  on  CE  as  a  side.     This  is  the  required  square. 

Prove. 

6.  Explain  how  to  draw  a  square  equal  to  a  given  polygon. 
MISCELLANEOUS   PROBLEMS   AND   EXERCISES 

|502.  Solve  the  following  problems  and  exercises : 

1.  Bisect  a  parallelogram  by  a  line  drawn  through  a  point 
on  its  perimeter. 

2.  Construct  an  equilateral  triangle  equivalent  to  a  given 
triangle. 

1.  Transform  the  given  triangle  into  an  equal  triangle  having 
one  angle  60°. 

2.  To  determine  the  length  of  the  side  of  the  equilateral  triangle, 
apply  the  theorem — two  triangles  having  an  angle  in  each  equal 
are  to  each  other  as  the  products  of  the  sides  including  the  equal 
angles. 

3.  The  base  of  a  triangle  is  18  feet.  Find  the  length  of  a  line 
parallel  to  the  base  which  bisects  the  triangle. 

4.  A  line  parallel  to  the  base  of  a  triangle  cuts  off  a  triangle 
equal  to  f  of  it.  If  one  side  of  the  triangle  is  12,  how  far  from 
the  vertex  does  the  line  cut  it  ? 

5.  Draw  through  a  vertex  of  a  triangle  lines  dividing  it: 

(1)  Into  two  parts  one  of  which  shall  be  (a)  J,  (6)  ^,  (c)  |  of 
the  other. 

(2)  Into  three  parts  in  the 
ratio  of  2:3:4. 

6.  To  bisect  the  surface  of 
a  triangle  by  a  Hne  through  a 
given  point  P  on  the  perimeter 
not  at  the  vertex  of  an  angle  Fig.  408 
(see  Fig.  408). 

Draw  the  median  BM,  also  PM,  BDJ  PM,  and  PD. 

Then,  APMD  =  APMB.        Why  ? 

.-.     AADP  =  AAMB  =  ABMC  =  PDCB.        Why? 


338  SECOND-YEAR  MATHEMATICS 

7.  The  sides  of  a  triangle  are  17,  10,  and  9.  The  altitude 
of  a  similar  triangle  upon  the  side  homologous  to  the  side  10  in 
the  given  triangle  is  14|-.  Find  all  the  sides  of  the  second  tri- 
angle. 

8.  The  side  of  a  square  (or  of  any  polygon,  or  the  radius  of  a 
circle)  is  a.  Find  the  side  (or  radius)  of  a  similar  figure  k  times 
as  large. 

9.  The  radii  of  two  circles  are  25  and  24.  Find  the  radius  of 
a  circle  equivalent  to  their  difference. 

10.  The  area  of  one  of  three  circles  is  equal  to  the  sum  of  the 
other  two,  and  their  radii  are  x,  x  —  7,  x-\-l.     Find  x. 

11.  The  difference  of  two  circles  whose  diameters  are  x-}-2 
and  X  is  equivalent  to  a  circle  whose  diameter  is  a:— 7.     Find  x. 

12.  The  area  of  a  rectangle  is  60  and  diagonal  is  13.  Find  its 
dimensions. 

13.  The  perimeter  of  a  rectangle  is  46  and  the  area  is  120. 
Find  its  dimensions. 

14.  The  perimeter  of  a  rectangle  is  62  and  the  diagonal  is 
25.     Find  its  area. 

16.  The  altitude  and  base  of  a  rectangle  are  in  the  ratio 
of  8  to  15  and  the  diagonal  is  34  feet.     Find  the  area. 

16.  The  dimensions  of  a  rectangle  are  in  the  ratio  of  2ah 
to  a^—b"^,  and  the  diagonal  is  aV+6V.     Find  the  area. 

17.  Compute  the  altitude  upon  the  hypotenuse  of  the  right 
triangle  ABC  in  terms  of  the  sides  of  the  right  angle. 

18.  The  diagonals  of  a  rhombus  are  2x  — 14  and  2x,  and  a 
sideisx+1.    Find  a;. 

19.  The  homologous  sides  of  two  similar  hexagons  are  9  in. 
and  12  in.  respectively.  Find  the  homologous  side  of  a 
similar  hexagon  (1)  equal  to  their  sum;  (2)  equal  to  their 
difference. 


POLYGONS.    CIRCLES.    PROPORTIONALITY      339 

Summary 

503.  The  following  theorems  have  been  proved  in 
the'  chapter. 

1.  The  area  of  a  rectangle  is  equal  to  the  product  of  the 
base  and  the  altitude. 

2.  The  area  of  a  parallelogram  is  equal  to  the  product 
of  the  base  and  the  altitude, 

3.  The  area  of  a  trapezoid  is  equal  to  one-half  the  prod- 
uct of  the  altitude  by  the  sum  of  the  bases. 

4.  The  area  of  a  regular  inscribed  polygon  is  equal  to  the 
product  of  one-half  the  perimeter  and  the  apothem. 

6.  The  area  of  a  regular  circumscribed  polygon  is  equal 
to  the  product  of  one-half  the  perimeter  and  the  radius. 

6.  The  area  of  a  circle  is  one-half  the  product  of  the  length 
of  the  circle  and  the  radius,  i.e.,  A  =  2CT. 

7.  The  area  of  a  circle  is  given  by  the  formula  A  =  Trr^. 

8.  The  area  of  a  sector  is  given  by  the  formula  A  =  ^a'r. 

9.  The  area  of  a  segment  of  a  circle  is  given  by  the 


formulas:  A  =  ^a'r—^a  \T^^  —  ^f 


or 


A  =  la'r-lr'  sin  X. 

10.  Two  parallelograms  are  to  each  other  as  the  products 
of  the  bases  and  altitudes. 

11.  Two  parallelograms  having  equal  bases  are  to  each 
other  as  the  altitudes. 

12.  Two  triangles  are  to  each  other  as  the  products  of  the 
bases  and  altitudes. 


340  SECOND-YEAR  MATHEMATICS 

13.  Areas  of  triangles  having  equal  bases  (altitudes)  are 
to  each  other  as  the  altitudes  (bases). 

14.  The  areas  of  two  triangles  having  an  angle  in  one 
equal  to  an  angle  in  the  other  are  to  each  other  as  the  products 
of  the  sides  including  the  equal  angles. 

15.  The  areas  of  similar  triangles  are  to  each  other  as  the 
squares  of  the  homologous  sides. 

16.  The  areas  of  similar  polygons  are  to  each  other  as  the 
squares  of  the  homologous  sides. 

17.  The  areas  of  two  circles  are  to  each  other  as  the  squares 
of  the  radii. 

504.  The  following  problems  of  construction  were 
taught : 

1.  Construct  a  square  equal  to  the  sum  of  two  or 
more  given  squares. 

2.  Construct  the  square  root  of  an  integral  number. 

3.  Transform  a  polygon  into  a  triangle. 

4.  Draw  a  square  equal  to  a  given  triangle. 


TABLE  OF  SINES,  COSINES,  AND  TANGENTS  OF 
ANGLES  FROM  1^-89° 


Angle 

Sine     C 

osine   Ta 

ngent 

Angle 

Sine     C 

osine 

Tangent 

1° 

0175 

9998 

0175 

460 

7193 

6947 

1-0355 

2 

0349 

9994 

0349 

47 

7314 

6820 

0724 

3 

0523 

9986 

0524 

48 

7431 

6691 

1106 

4 

0698    . 

9976 

0699 

49 

7547 

6561 

1504 

5 

0872 

9962 

0875 

50 

7660 

6428 

1918 

6 

1045 

9945 

1051 

51 

'^iv- 

6293 

2349 

7 

1219 

9925 

1228 

52 

7880 

6157 

2799 

8 

1392 

9903 

1405 

53 

7986, 

6018 

3270 

9 

1564 

9877 

1584 

54 

8090 

5878 

3764 

10 

1736 

9848 

1763 

55 

8192 

5736 

4281 

II 

1908 

9816 

1944 

56 

8290 

5592 

4826 

12 

2079 

9781 

2126 

57 

8387 

5446 

5399 

13 

2250 

9744 

2309 

58 

8480 

5299 

6003 

14 

2419 

9703 

2493 

59 

8572 

5150 

6643 

15 

2588 

9659 

2679 

60 

8660 

5000 

7321 

i6 

2756 

9613 

2867 

61 

8746 

4848 

8040 

17 

2924 

9563 

3057 

62 

8829 

4695 

8807 

i8 

3090 

9511 

3249 

63 

8910 

4540 

9626 

19 

3256 

9455 

3443 

64 

8988 

4384 

2 

0503 

20 

3420 

9397 

3640 

65 

9063 

4226 

2 

1445 

21 

3584 

9336 

3839 

66 

9135 

4067 

2 

2460 

22 

3746 

9272 

4040 

^A 

9205 

3907 

2 

3559 

23 

3907 

9205 

4245 

68 

9272 

3746 

2 

4751 

24 

4067 

9135 

4452 

69 

9336 

3584 

2 

6051 

25 

4226 

9063 

4663 

70 

9397 

3420 

2 

7475 

26 

4384 

8988 

4877 

71 

9455 

3256 

2 

9042 

27 

4540 

8910 

5095 

72 

9511 

3090 

3 

0777 

28 

4695 

8829 

5317 

73 

9563 

2924 

3 

2709 

29 

4848 

8746 

5543 

74 

9613 

2756 

3 

4874 

30 

5000 

8660 

5774 

75 

9659 

2588 

3 

7321 

31 

5150 

8572 

6009 

76 

9703 

2419 

4 

0108 

32 

5299 

8480 

6249 

77 

9744 

2250 

4 

3315 

33 

5446 

8387 

6494 

78 

9781 

2079 

4 

7046 

34 

5592 

8290 

6745 

79 

9816 

1908 

5 

1446 

35 

5736 

8192 

7CXD2 

80 

9848 

1736 

5 

6713 

36 

5878 

8090 

7265 

81 

9877 

1564 

6 

3138 

37 

6018 

7986 

7536 

82 

9903 

1392 

7 

1154 

38 

6157 

7880 

7813 

83 

9925 

1219 

8 

1443 

39 

6293 

7771 

8098 

84 

9945 

1045 

9 

5144 

40 

6428 

.7660 

8391 

85 

.9962 

0872 

II 

.4301 

41 

.6561 

•7547 

.8693 

86 

.9976 

0698 

14 

.3006 

42 

.6691 

•7431 

.9004 

87 

.9986 

•0523 

19 

.0811 

43 

.6820 

•7314 

•9325 

88 

•9994 

0349 

28 

•6363 

44 

.6947 

•7193 

•9657 

89 

.9998 

0175 

57 

.2900 

45 

.7071 

.7071   I 

.0000 

341 


TABLE  OF  POWERS  AND  ROOTS 

No. 

Squares 

Cubes 

Square 
Roots 

Cube 
Roots 

No. 

Squares 

Cubes 

Square 
Roots 

Cube 
Roots 

I 

I 

I 

1. 000 

1. 000 

51 

2,601 

132,651 

7. 141 

3.708 

2 

4 

8 

1-414 

1-259 

52 

2,704 

140,608 

211 

3 

732 

3 

9 

27 

1.732 

1.442 

SZ 

2,809 

148,877 

280 

3 

756 

4 

16 

64 

2.000 

1-587 

54 

2,916 

157,464 

-7 

348 

3 

779 

5 

25 

125 

2.236 

1.709 

55 

3,025 

166,375 

416 

3 

802 

6 

36 

216 

2.449 

1. 817 

56 

3,136 

175,616 

483 

3 

82s 

7 

49 

343 

2.64s 

1. 912 

57 

3,249 

185,193 

549 

3 

848 

8 

64 

512 

2.828 

2.000 

58 

3,364 

195,112 

615 

3 

870 

9 

81 

729 

3.000 

2.080 

59 

3,481 

205,379 

681 

3 

892 

10 

II 

100 

1,000 

3.162 

2.154 

60 

3,600 

216,000 

745 

_3_ 
3 

914 

121 

1,331 

3-316 

2.223 

61 

3,721 

226,981 

810 

936 

12 

144 

1,728 

3  464 

2.289 

62 

3,844 

238,328 

874 

3 

957 

13 

169 

2,197 

3  605 

2-351 

63 

3,969 

250,047 

937 

3 

979 

14 

196 

2,744 

3-741 

2.410 

64 

4,096 

262,144 

8 

000 

4 

000 

15 

225 

3,375 

3.872 

2.466 

65 

4,225 

274,625 

8 

062 

4 

020 

i6 

256 

4,096 

4.000 

2.519 

66 

4,356 

287,496 

8 

124 

4 

041 

17 

289 

4,913 

4-123 

2.571 

67 

4,489 

300,763 

8 

185 

4 

061 

i8 

324 

5,832 

4.242 

2.620 

68 

4,624 

314,432 

8 

246 

4 

081 

19 

361 

6,859 

4.358 

2.668 

69 
70 

4,761 

328,509 

8 

306 

4 

lOI 

20 

400 

8,000 

4.472 

2.714 

4,900 

343,000 

8 

366 

4 

121 

21 

441 

9,261 

4-582 

2-758 

71 

5,041 

357,911 

8 

426 

4 

140 

22 

484 

10,648 

4.690 

2.802 

72 

5,184 

373,248 

8 

48s 

4 

160 

23 

529 

12,167 

4-795 

2.843 

73 

5,329 

389,017 

8 

544 

4 

179 

24 

576 

13,824 

4.898 

2.884 

74 

5,476 

405,224 

8 

602 

4 

198 

25 

625 

15,625 

5.000 

2.924 

75 

5,625 

421,87s 

8 

660 

4 

217 

26 

676 

17,576 

5-099 

2.962 

76 

5,776 

438,976 

8 

717 

4 

235 

27 

729 

19,683 

5.196 

3.000 

77 

5,929 

456,533 

8 

774 

4 

254 

28 

784 

21,952 

5-291 

3-036 

78 

6,084 

474,552 

8 

831 

4 

272 

29 

841 

24,389 

5-385 

3.072 

79 

6,241 

493,039 

8 

888 

4 

290 

30 

900 

27,000 

5-477 

3.107 

80 

6,400 

512,000 

8 

944 

4 

308 

31 

961 

29,791 

5-567 

3 -141 

81 

6,561 

531,441 

9 

000 

4 

326 

32 

1,024 

32,768 

5-656 

3-174 

82 

6,724 

551,368 

9 

05s 

4 

344 

33 

1,089 

35,937 

5-744 

3.207 

83 

6,889 

571,787 

9 

no 

4 

362 

34 

1,156 

39,304 

5-830 

3-239 

84 

7.056 

592,704 

9 

165 

4 

379 

35 

1,225 

42,875 

5.916 

3.271 

85 

7,225 

614,125 

9 

219 

4 

396 

36 

1,296 

46,656 

6.000 

3-301 

86 

7,396 

636,056 

9 

273 

4 

414 

?,7 

I, -369 

50,653 

6.082 

3-332 

87 

7,569 

658,503 

9 

327 

4 

431 

3B 

1,444 

54,872 

6.164 

3-361 

88 

7,744 

681,472 

9 

380 

4 

447 

39 

1,521 

59,319 

6.244 

3-391 

89 

7,921 

704,969 

9 

433 

4 

464 

40 

1,600 

64,000 

6.324 

3-419 

90 

8,100 

729,000 

9 

486 

4 

481 

41 

1,681 

68,921 

6.403 

3  448 

91 

8,281 

753,571 

9 

539 

4 

497 

42 

1,764 

74,088 

6.480 

3  476 

92 

8,464 

778,688 

9 

591 

4 

514 

43 

1,849 

79,507 

6.557 

3-503 

93 

8,649 

804,357 

9 

643 

4 

530 

44 

1,936 

85,184 

6.633 

3-530 

94 

8,836 

830,584 

9 

695 

4 

546 

45 

2,025 

91,125 

6.708 

3-556 

95 

9,025 

857,375 

9 

746 

4 

562 

46 

2,116 

97,336 

6.782 

3-583 

96 

9,216 

884,736 

9 

797 

4 

578 

47 

2,209 

103,823 

6-855 

3-608 

97 

9,409 

912,673 

9 

848 

4 

594 

48 

2,304 

110,592 

6.928 

3-634 

98 

9,604 

941,192 

9 

899 

4 

610 

49 

2,401 

117,649 

7.000 

3  659 

99 
100 

9,801 

970,299 

9 

949 

4 

626 

50 

2,500 

125,000 

7.071 

3.684 

10,000 

1 ,000,000 

10 

000 

4 

641 

SYMBOLS 

= 

equals,  equals,  is  equal  to 

(D 

circles 

> 

is  greater  than 

.'. 

hence,  therefore. 

< 

is  less  than 

'.' 

since 

II 

parallel,  is  parallel  to 

= 

identical,  is  identical  to 

± 

perpendicular,  is  perpen- 

= 

approaches 

dicular  to 

f 

plus 

oo 

similar,  is  similar  to 

_ 

minus 

CO 

congruent,  is  congruent  to 

7^ 

does  not  equal 

z 

angle 

rtZ 

right  angle 

A 

angles 

A 

triangle 

D 

parallelogram 

A 

triangles 

□ 

rectangle 

,-^ 

arc 

O 

circle 

FORMULAS 

a2 4-62=^2;  relation  between  the  sides  of  a  right  triangle. 
a2-f62±2a6'  =  c2:   relations  between   the  sides  of  an   oblique 

triangle. 
c  =  2TTr=Trd:  length  of  a  circle. 
h'h:  area  of  a  parallelogram,  of  rectangle. 
0?:  area  of  a  square. 
\hh,  \ah  sin  C,  \r{a-Vh^c),  ^ ,  V s{s-a){s-h){s-c): 

area  of  a  triangle. 
\a?V^:  area  of  an  equilateral  triangle. 
hm=\h(][)-^h') :  area  of  a  trapezoid. 
^p-a:  area  of  a  regular  inscribed  polygon, 
^p  •  r:  area  of  a  regular  circumscribed  polygon.  ^^ 

§cr='7rr^:  area  of  a  circle. 

sin  A  =  -  ,  cos  A  =  - ,  tan  ^  =  r  ,  sin^A+cos^A  =  1 ,  tan  A  =  ^^^  ^  . 

343 


[References  are 

Addition  and  subtraction, 
195- 

Ahmes .  . , 204, 

Algebraic  method 

Alternation 

Altitude  of  triangle 

Analysis:  method  of 

proof  by 

Angle:  central 

inscribed 

measure  of 297, 

Angles:  exterior 

interior 

of  parallelogram ...  120, 

Antecedent 

Antipho 

Apollonius 

Apothem 

Approximate       measure- 
ment   

Arc :  intercepted 

major,  minor 

subtended 

Arch,  Gothic 

Archimedes.  .160, 238,  290, 

Archytas 204, 

Areas:  comparison  of . . . . 

of  circle 

of  parallelogram 

of  rectangle 


INDEX 

to  sections,  not  to  pages] 

of  triangle 465 

197  of     regular     inscribed 

451  polygon 484 

90  of  reg  ular  circumscribed 

192             polygon 485 

470  of  segment  of  circle. ...  491 

112  of  sector  of  circle.. 490 

113  Assumptions,  list  of .  .  .  .  1-79 
297 

295      Cardan 238 

298      Center  of  gravity 415 

89      Central  angle 297 

88      Chord 279 

121          subtending 282 

194      Circle:  length  of 450 

452  sector  of 489 

290         segment  of 299 

484      Circles:  tangent 288 

axis  of 390 

257          poles  of 390 

294      Circular  motion 302 

276      Circumcentcr 420 

282  Circumscribed  polygon. . .  433 

274  Commensurable,     magni- 

451          tudes 164 

249      Comparison 101 

459  Compasses,  proportional.  158 

489      Complex  fractions 328 

482      Concentric  circles 275 

481      Conclusion 80 

344 


INDEX 


345 


[References  are  to  sections,  not  to  pages] 


Concurrent  lines 416 

Conditions:  for  congruent 

triangles 215 

for  similar  triangles. ...  216 
Congruent  triangles,  con- 
ditions for 215 

Consequent 194 

Converse:  of  a  theorem..   114 

proof  of 115 

Cosine  of  angle 248 

Diagonal  scale 158 

Diedral  angle 144,  380 

plane  angle  of 146 

size  of 145 

Diedral  angles,  classifica- 
tion of 147 

Direct  variation 200 

Distance:  polar 393 

spherical 391 

Division  of  line-segments: 

external 171, 176 

harmonic 173 

internal 176 

in  mean  and  extreme 
ratio 316 

Eclipse,  lunar 303 

EUmination 97 

by  addition  or  subtrac- 
tion       98 

Equations:  hteral 464 

in  two  unknowns 97 

quadratic 131 

EucUd,  Elements  of     204, 238 
Eudoxus 111,204,452 


Euler 426 

Excenter 425 

Exterior  angles,  sum  of . .     89 
Factoring.  .  .  189-190, 472-478 

Fallacies,  geometrical 77 

Fermat,  Pierre  de 234 

Feuerbach 426 

Fourth  proportional J^ 

Fractions 320-328 

complex 328 

Fractional  equations 329 

Functions,  trigonometric.  251 

Gauss,  Carl  Friedrich.. .  .  444 

Geometrical  fallacies ....  77 

Gothic  arch 274 

Graphical  method  of  solu- 
tion   99 

Gravity,  center  of 415 

Great  circle  of  sphere. .  .  .  390 

Harmonic  divisions 173 

Heron  of  Alexandria    238, 290 

Hippocrates  of  Chios,  111,  452 

Historical  notes,    85,  100,  111 

112,  161,  163,  172,  234,  238, 

248,  276,  290,  299,  301,  364, 

426,  444,  452,  463,  469 

Homologous  parts 213 

Hypothesis 80 

Incenter 424 

Incommensurable,  case. .  166 

magnitudes 165 

Inequahties,  axioms  of. .  338 

problems  of 339 


346 


SECOND-YEAR  MATHEMATICS 


[References  are  to  sections,  not  to  pages] 


Indirect  method  of  proof.  109 

Inductive  method 87 

Inscribed  angle 295 

measurement  of 298 

Inscribed    polygon 432 

Interest  problems 332 

Interior  angles,  sum  of . .  88 

Inversion 193 

Irrational  numbers 234 

Isosceles  trapezoid ......  128 

Kite 133 

Klein,  Felix p.  v 

Length  of  circle 450 

Lindeman 451 

Lines:  concurrent 416 

perpendicular  to  plane.  177 

relative  positions  of 140 

Line-segment:        divided 

harmonically 173 

divided  into  mean  and 

extreme  ratio 316 

measurement  of 155 

ratio  of 156 

Line-segments,       propor- 
tional in  circles 313 

Literal  equations 464 

Locus 135,404 

determination  of 405 

proof  of 406 

Logic 8 

Lunar  eclipse 303 

Mathematical    forms    in 

architecture,  pp.  74,  160, 
245,  280,  323 

puzzle 299 


Mean  and  extreme  ratio . .  316 
Mean  proportional .  . .  227, 231 
Measurement:    of  angles, 

294, 297 

of  inscribed  angles 298 

Median 414 

of  trapezoid 161 

Methods:  of  proof,  71,  81,  82, 

87,  90,  91,  109,  112 

of  elimination 97 

Motion,  circular 302 

Orthocenter 423 

Pantograph 158 

Pappus 290,463 

Parallelogram 105, 128 

construction  of 107 

properties  of 117 

uses  of 106 

Percentage  problems 332 

Perpendicular  planes 148 

Pi  (^) 450 

Plane:  determination  of..   139 

representation  of 142 

tangent  to  sphere 398 

Planes:  perpendicular 148 

perpendicular  to  line. . .   177 

in  space 364 

Plato 112,204,238 

Point  of  trisection 418 

Polar  distance 393 

Polygon:  circumscribed..  433 

inscribed.. 432 

regular 430 

similar 209 

Proclus 276 


INDEX 


347 


[References  are  to  sections,  not  to  pages] 


Projection:  of  a  point  225,354 

of  a  segment 226 

of  a  solid 353 

Proportion 157 

Proportional:  compasses.   158 

fom-th 174 

line-segments 158 

mean 227 

third 175 

Proportionality:  of  areas  492 
test  of 186 

Prism 136 

bases  of 137 

lateral  face 137 

lateral  surface  of 137 

Proof:    general  directions 

for 80 

methods  of 79,81,82 

need  of 78 

Properties    of    parallelo- 
gram    117 

Puzzle,  mathematical. . . .  299 

Pythagoras,  theorem  of, 

233,  462 

Quadrant 394 

Quadratic  equations, 

131,  235-237 

in  two  unknowns 264 

solved  by  gi'aph 265 

Quadrilaterals 128 

Quotient,      found     from 
graph 211 

Radical 228 

simplification  of 229 


Ratio  of  Hne-segments .. .  156 

trigonometric 248 

RationaHzing  the  denomi- 
nator    254 

Rectangle 128 

Reductio  ad  absurdum. ...  Ill 

Reasoning 76 

Reciprocal 327 

Regular  polygon 430 

Relations,  trigonometric.  334 

Reviews 86 

Rhomboid 128 

Rhombus 128 

Scale,  diagonal 158 

Secant  of  circle 277 

Semicircle 276 

Similar  polygons 209 

Sine  of  angle 248 

Sphere 386 

center,  radius,  diameter  386 

great  circle  of 390 

section  of 388 

Spherical  distance 391 

Square 128 

Study  helps p.  xix 

Substitution 91,100 

Subtending   chord 282 

Superposition 81 

Siuface,  prismatic 136 

Symbols p.  343 

Symmetry 134 

Tangent:  Une,  plane.  .  .  .  398 

of  angle 248 

circles 288 

to  circle 277 


348 


SECOND-YEAR  MATHEMATICS 


[References  are  to  sections,  not  to  pages] 


Test  of  proportionality. . .  186 

Thales 204 

Third  proportional 175 

Trapezoid 128,  132 

median  of. 161 

Triangle:  area  of 465 

altitude  of 470 

Trigonometric :  functions .  251 

relations 334 


Trisection  point 418 

Units  of  angular  measure  294 

Variation:  direct 201 

inverse 202 

Weight  problems 333 

Work  problems 102 


<0 


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